(a) We would like to solve $color{#4257b2}5^{x}=625$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}625=5^{4}$, so we can replace $color{#4257b2}625$ from the right side by $color{#4257b2}5^{4}$.

$$
5^{x}=625
$$

$$
5^{x}=5^{4}
$$

Now the two sides have the same base $color{#4257b2}5$, so the exponent of the left side will equal the exponent of the right side.

$$
5^{x}=5^{4}
$$

$$
x=4
$$

So the solution of the equation is $boxed{ x=4 }$

(b) We would like to solve $color{#4257b2}4^{2x}=2^{5-x}$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}4=2^{2}$, so we can replace $color{#4257b2}4$ from the left side by $color{#4257b2}2^{2}$.

$$
4^{2x}=2^{5-x}
$$

$$
(2^{2})^{2x}=2^{5-x}
$$

But we know from the exponent property that $color{#4257b2}(x^{a})^{b}=x^{a b}$, so we can use this property to simplify the left side.

$$
2^{(2)cdot (2x)}=2^{5-x}
$$

$$
2^{4x}=2^{5-x}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
4x=5-x
$$

Now we can add $color{#4257b2}x$ to each side to make the variable $color{#4257b2}x$ in the left side alone.

$$
4x+x=5-x+x
$$

$$
5x=5
$$

Now we can divide the two sides by $color{#4257b2}5$

$$
dfrac{5x}{5}=dfrac{5}{5}
$$

$$
x=1
$$

So the solution of the equation is $boxed{ x=1 }$

(c) We would like to solve $color{#4257b2}9^{x+1}=27^{2x-3}$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}9=3^{2}$ and $color{#4257b2}27=3^{3}$, so we can replace $color{#4257b2}9$ and $color{#4257b2}27$ from our equation by $color{#4257b2}3^{2}$ and $color{#4257b2}3^{3}$.

$$
9^{x+1}=27^{2x-3}
$$

$$
(3^{2})^{x+1}=(3^{3})^{2x-3}
$$

But we know from the exponent property that $color{#4257b2}(x^{a})^{b}=x^{a b}$, so we can use this property to simplify our equation.

$$
3^{(2)cdot (x+1)}=3^{(3)cdot (2x-3)}
$$

$$
3^{2x+2}=3^{6x-9}
$$

Now the two sides have the same base $color{#4257b2}3$, so the exponent of the left side will equal the exponent of the right side.

$$
2x+2=6x-9
$$

Now we can add $color{#4257b2}9-2x$ to each side to make the variable $color{#4257b2}x$ in the right side alone.

$$
2x+2=6x-9
$$

$$
2x+2+9-2x=6x-9+9-2x
$$

$$
11=4x
$$

Now we can divide the two sides by $color{#4257b2}4$

$$
dfrac{11}{4}=dfrac{4x}{4x}
$$

$$
dfrac{11}{4}=x
$$

So the solution of the equation is $boxed{ x=dfrac{11}{4} }$

(d) We would like to solve $color{#4257b2}8^{x-1}=sqrt[3]{16}$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}8=2^{3}$ and $color{#4257b2}16=2^{4}$, so we can replace $color{#4257b2}8$ and $color{#4257b2}16$ from our equation by $color{#4257b2}2^{3}$ and $color{#4257b2}2^{4}$.

$$
8^{x-1}=sqrt[3]{2^{4}}
$$

$$
(2^{3})^{x-1}=sqrt[3]{2^{4}}
$$

Also we know that $color{#4257b2}sqrt[a]{x^{b}}=x^{frac{b}{a}}$, so we can use this fact to replace $color{#4257b2}sqrt[3]{2^{4}}$ from the right side by $color{#4257b2}2^{frac{4}{3}}$ to simplify our equation.

$$
(2^{3})^{x-1}=sqrt[3]{2^{4}}
$$

$$
(2^{3})^{x-1}=2^{frac{4}{3}}
$$

But we know from the exponent property that $color{#4257b2}(x^{a})^{b}=x^{a b}$, so we can use this property to simplify the left side.

$$
2^{(3)cdot (x-1)}=2^{frac{4}{3}}
$$

$$
2^{3x-3}=2^{frac{4}{3}}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
3x-3=dfrac{4}{3}
$$

Now we can add $color{#4257b2}3$ to each side to make the variable $color{#4257b2}x$ in the left side alone.

$$
3x-3+3=dfrac{4}{3}+3
$$

$$
3x=dfrac{13}{3}
$$

Now we can divide the two sides by $color{#4257b2}3$

$$
dfrac{3x}{3}=dfrac{dfrac{13}{3}}{3}
$$

$$
x=dfrac{13}{9}
$$

So the solution of the equation is $boxed{ x=dfrac{13}{9} }$

(e) We would like to solve $color{#4257b2}2^{3x}=dfrac{1}{2}$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others, so the first step is to use the property of exponent where $color{#4257b2}dfrac{1}{x^{a}}=x^{-a}$ to replace $color{#4257b2}dfrac{1}{2}$ from the right side by $color{#4257b2}2^{-1}$.

$$
2^{3x}=dfrac{1}{2}
$$

$$
2^{3x}=2^{-1}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
2^{3x}=2^{-1}
$$

$$
3x=-1
$$

Now we can divide the two sides by $color{#4257b2}3$.

$$
dfrac{3x}{3}=dfrac{-1}{3}
$$

$$
x=-dfrac{1}{3}
$$

So the solution of the equation is $boxed{ x=-dfrac{1}{3} }$

(f) We would like to solve $color{#4257b2}4^{2x}=dfrac{1}{16}$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}16=4^{2}$, so we can replace $color{#4257b2}16$ from the right side by $color{#4257b2}4^{2}$.

$$
4^{2x}=dfrac{1}{16}
$$

$$
4^{2x}=dfrac{1}{4^{2}}
$$

Now we can use the property of exponent where $color{#4257b2}dfrac{1}{x^{a}}=x^{-a}$ to replace $color{#4257b2}dfrac{1}{4^{2}}$ from the right side by $color{#4257b2}4^{-2}$.

$$
4^{2x}=dfrac{1}{4^{2}}
$$

$$
4^{2x}=4^{-2}
$$

Now the two sides have the same base $color{#4257b2}4$, so the exponent of the left side will equal the exponent of the right side.

$$
4^{2x}=4^{-2}
$$

$$
2x=-2
$$

Now we can divide the two sides by $color{#4257b2}2$.

$$
dfrac{2x}{2}=dfrac{-2}{2}
$$

$$
x=-1
$$

So the solution of the equation is $boxed{ x=-1 }$

Large{$text{$text{color{#c34632}(a) $x=4$ (c) $x=dfrac{11}{4}$ (e) $-dfrac{1}{2}$
\
\
\
Large{color{#c34632}(b) $x=1$ (d) $x=dfrac{13}{9}$ (f) $-1$}$}$

(a) We would like to solve $color{#4257b2}2^{x}=17$. First, we note that the two sides have different bases, so we can take $color{#4257b2}log_{2}$ for each side to simplify the left side.

$$
2^{x}=17
$$

$$
log_{2} 2^{x}=log_{2} 17
$$

Now we note that the left side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log_{2} 2^{x}=log_{2} 17
$$

$$
x log_{2} 2=log_{2} 17
$$

Now we can use the property of logarithm $color{#4257b2}log_{a} a=1$, to replace $color{#4257b2}log_{2} 2$ by $color{#4257b2}1$.

$$
xcdot (1)=log_{2} 17
$$

$$
x=log_{2} 17
$$

Now we can use the calculator to determine $color{#4257b2}log_{2} 17$ to find the value of $color{#4257b2}x$.

$$
x=4.087
$$

So the solution of the equation is $boxed{ x=4.087 }$

(b) We would like to solve $color{#4257b2}6^{x}=231$. First, we note that the two sides have different bases, so we can take $color{#4257b2}log_{6}$ for each side to simplify the left side.

$$
6^{x}=231
$$

$$
log_{6} 6^{x}=log_{6} 231
$$

Now we note that the left side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log_{6} 6^{x}=log_{6} 231
$$

$$
x log_{6} 6=log_{6} 231
$$

Now we can use the property of logarithm $color{#4257b2}log_{a} a=1$, to replace $color{#4257b2}log_{6} 6$ by $color{#4257b2}1$.

$$
xcdot (1)=log_{6} 231
$$

$$
x=log_{6} 231
$$

Now we can use the calculator to determine $color{#4257b2}log_{6} 231$ to find the value of $color{#4257b2}x$.

$$
x=3.037
$$

So the solution of the equation is $boxed{ x=3.037 }$

(c) We would like to solve $color{#4257b2}30(5^{x})=150$. First, we can divide the two sides by $color{#4257b2}30$ to simplify.

$$
30(5^{x})=150
$$

$$
dfrac{30(5^{x})}{30}=dfrac{150}{30}
$$

$$
5^{x}=5
$$

Now we note that the two sides have the same base $color{#4257b2}5$, so the exponent of the left side must equal the exponent of the right side.

$$
5^{x}=5
$$

$$
5^{x}=5^{1}
$$

$$
x=1
$$

So the solution of the equation is $boxed{ x=1 }$

(d) We would like to solve $color{#4257b2}210=40 (1.5)^{x}$. First, we can divide the two sides by $color{#4257b2}40$ to simplify.

$$
210=40 (1.5)^{x}
$$

$$
dfrac{210}{40}=dfrac{40 (1.5)^{x}}{40}
$$

$$
5.25=(1.5)^{x}
$$

Now we note that the two sides have different bases, so we can take $color{#4257b2}log_{1.5}$ for each side to simplify the right side.

$$
log_{1.5} 5.25=log_{1.5} (1.5)^{x}
$$

Now we note that the right side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log_{1.5} 5.25=x log_{1.5} 1.5
$$

Now we can use the property of logarithm $color{#4257b2}log_{a} a=1$, to replace $color{#4257b2}log_{1.5} 1.5$ by $color{#4257b2}1$.

$$
log_{1.5} 5.25=xcdot (1)
$$

$$
log_{1.5} 5.25=x
$$

Now we can use the calculator to determine $color{#4257b2}log_{1.5} 5.25$ to find the value of $color{#4257b2}x$.

$$
4.09=x
$$

So the solution of the equation is $boxed{ x=4.09 }$

(e) We would like to solve $color{#4257b2}5^{1-x}=10$. First, we note that the two sides have different bases, so we can take $color{#4257b2}log_{5}$ for each side to simplify the left side.

$$
5^{1-x}=10
$$

$$
log_{5} 5^{1-x}=log_{5} 10
$$

Now we note that the left side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
(1-x) log_{5} 5=log_{5} 10
$$

Now we can use the property of logarithm $color{#4257b2}log_{a} a=1$, to replace $color{#4257b2}log_{5} 5$ by $color{#4257b2}1$.

$$
(1-x)cdot (1)=log_{5} 10
$$

$$
1-x=log_{5} 10
$$

Now we can add $color{#4257b2}x-log_{5} 10$ to each side to make variable $color{#4257b2}x$ in the right side alone.

$$
1-x+x-log_{5} 10=log_{5} 10+x-log_{5} 10
$$

$$
1-log_{5} 10=x
$$

Now we can use the calculator to determine $color{#4257b2}log_{5} 10$ to find the value of $color{#4257b2}x$.

$$
1-1.431=x
$$

$$
-0.431=x
$$

So the solution of the equation is $boxed{ x=-0.431 }$

(f) We would like to solve $color{#4257b2}6^{frac{x}{3}}=30$. First, we note that the two sides have different bases, so we can take $color{#4257b2}log_{6}$ for each side to simplify the left side.

$$
6^{frac{x}{3}}=30
$$

$$
log_{6} 6^{frac{x}{3}}=log_{6} 30
$$

Now we note that the left side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
left(dfrac{x}{3}right) log_{6} 6=log_{6} 30
$$

Now we can use the property of logarithm $color{#4257b2}log_{a} a=1$, to replace $color{#4257b2}log_{6} 6$ by $color{#4257b2}1$.

$$
left(dfrac{x}{3}right)cdot (1)=log_{6} 30
$$

$$
dfrac{x}{3}=log_{6} 30
$$

Now we can multiply the two sides by $color{#4257b2}3$ to make variable $color{#4257b2}x$ in the left side alone.

$$
(3)cdot left(dfrac{x}{3}right)=3 log_{6} 30
$$

$$
x=3log_{6} 30
$$

Now we can use the calculator to determine $color{#4257b2}log_{6} 30$ to find the value of $color{#4257b2}x$.

$$
x=3cdot (1.898)
$$

$$
x=5.695
$$

So the solution of the equation is $boxed{ x=5.695 }$

Large{$text{$text{color{#c34632}(a) $x=4.087$ (c) $x=1$ (e) $x=-0.431$
\
\
Large{color{#c34632}(b) $x=3.037$ (d) $x=4.09$ (f) $x=5.695$}$}$

(a) We would like to solve $color{#4257b2}x=log_{3} 243$. First, we need to make our equation on the exponential form, so we can convert from the logarithmic form to the exponential form where we know if $color{#4257b2}x=log_{a} b$ then $color{#4257b2}a^{x}=b$.

$$
x=log_{3} 243
$$

$$
3^{x}=243
$$

Now we converted our equation to the exponential form, so the next step is to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}243=3^{5}$, so we can replace $color{#4257b2}243$ from the right side by $color{#4257b2}3^{5}$.

$$
3^{x}=3^{5}
$$

Now the two sides have the same base $color{#4257b2}3$, so the exponent of the left side will equal the exponent of the right side.

$$
x=5
$$

So the solutions of the equation is $boxed{ x=5 }$

(b) We would like to solve $color{#4257b2}x=log_{6} 216$. First, we need to make our equation on the exponential form, so we can convert from the logarithmic form to the exponential form where we know if $color{#4257b2}x=log_{a} b$ then $color{#4257b2}a^{x}=b$.

$$
x=log_{6} 216
$$

$$
6^{x}=216
$$

Now we converted our equation to the exponential form, so the next step is to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}216=6^{3}$, so we can replace $color{#4257b2}216$ from the right side by $color{#4257b2}6^{3}$.

$$
6^{x}=6^{3}
$$

Now the two sides have the same base $color{#4257b2}6$, so the exponent of the left side will equal the exponent of the right side.

$$
x=3
$$

So the solutions of the equation is $boxed{ x=3 }$

(c) We would like to solve $color{#4257b2}x=log_{5} 5 sqrt{5}$. First, we need to make our equation on the exponential form, so we can convert from the logarithmic form to the exponential form where we know if $color{#4257b2}x=log_{a} b$ then $color{#4257b2}a^{x}=b$.

$$
x=log_{5} 5 sqrt{5}
$$

$$
5^{x}=5 sqrt{5}
$$

Now we converted our equation to the exponential form, so the next step is to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}sqrt{x}=x^{frac{1}{2}}$, so we can replace $color{#4257b2}sqrt{5}$ from the right side by $color{#4257b2}5^{frac{1}{2}}$.

$$
5^{x}=(5)cdot (5)^{frac{1}{2}}
$$

Now we can use the property of exponents where $color{#4257b2}x^{a}cdot x^{b}=x^{ab}$.

$$
5^{x}=5^{1+frac{1}{2}}
$$

$$
5^{x}=5^{frac{3}{2}}
$$

Now the two sides have the same base $color{#4257b2}5$, so the exponent of the left side will equal the exponent of the right side.

$$
x=dfrac{3}{2}
$$

So the solutions of the equation is $boxed{ x=dfrac{3}{2} }$

(d) We would like to solve $color{#4257b2}x=log_{2} sqrt[5]{8}$. First, we need to make our equation on the exponential form, so we can convert from the logarithmic form to the exponential form where we know if $color{#4257b2}x=log_{a} b$ then $color{#4257b2}a^{x}=b$.

$$
x=log_{2} sqrt[5]{8}
$$

$$
2^{x}=sqrt[5]{8}
$$

Now we converted our equation to the exponential form, so the next step is to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}8=2^{3}$, so we can replace $color{#4257b2}8$ from the right side by $color{#4257b2}2^{3}$ to simplify the right side.

$$
2^{x}=sqrt[5]{2^{3}}
$$

Now we know that $color{#4257b2}sqrt[a]{x^{b}}=x^{frac{b}{a}}$, so we can replace $color{#4257b2}sqrt[5]2^{3}$ from the right side by $color{#4257b2}2^{frac{3}{5}}$.

$$
2^{x}=2^{frac{3}{5}}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
x=dfrac{3}{5}
$$

So the solutions of the equation is $boxed{ x=dfrac{3}{5} }$

(e) We would like to solve $color{#4257b2}x=log_{2} left(dfrac{1}{4}right)$. First, we need to make our equation on the exponential form, so we can convert from the logarithmic form to the exponential form where we know if $color{#4257b2}x=log_{a} b$ then $color{#4257b2}a^{x}=b$.

$$
x=log_{2} left(dfrac{1}{4}right)
$$

$$
2^{x}=dfrac{1}{4}
$$

Now we converted our equation to the exponential form, so the next step is to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}4=2^{2}$, so we can replace $color{#4257b2}4$ from the right side by $color{#4257b2}2^{2}$.

$$
2^{x}=dfrac{1}{2^{2}}
$$

Now we can use the property of exponents where $color{#4257b2}dfrac{1}{x^{a}}=x^{-a}$.

$$
2^{x}=2^{-2}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
x=-2
$$

So the solutions of the equation is $boxed{ x=-2 }$

(f) We would like to solve $color{#4257b2}x=log_{3} left(dfrac{1}{sqrt{3}}right)$. First, we need to make our equation on the exponential form, so we can convert from the logarithmic form to the exponential form where we know if $color{#4257b2}x=log_{a} b$ then $color{#4257b2}a^{x}=b$.

$$
x=log_{3} left(dfrac{1}{sqrt{3}}right)
$$

$$
3^{x}=dfrac{1}{sqrt{3}}
$$

Now we converted our equation to the exponential form, so the next step is to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}sqrt{x}=x^{frac{1}{2}}$, so we can replace $color{#4257b2}sqrt{3}$ from the right side by $color{#4257b2}3^{frac{1}{2}}$.

$$
3^{x}=dfrac{1}{3^{frac{1}{2}}}
$$

Now we can use the property of exponents where $color{#4257b2}dfrac{1}{x^{a}}=x^{-a}$.

$$
3^{x}=(3)^{-frac{1}{2}}
$$

Now the two sides have the same base $color{#4257b2}3$, so the exponent of the left side will equal the exponent of the right side.

$$
x=-dfrac{1}{2}
$$

So the solutions of the equation is $boxed{ x=-dfrac{1}{2} }$

Large{$text{$text{color{#c34632}(a) $x=5$ (c) $x=dfrac{3}{2}$ (e) $-2$
\
\
\
Large{color{#c34632}(b) $x=3$ (d) $x=dfrac{3}{5}$ (f) $-dfrac{1}{2}$}$}$

We would like to know the time $color{#4257b2}t$ which the radioactive substance with mass $color{#4257b2}P=300 text{g}$ will take to decay to some values of masses if we know that the formula to calculate the mass is $color{#4257b2}M(t)=Pleft(dfrac{1}{2}right)^{frac{t}{h}}$ and we know that the half life time of the substance $color{#4257b2}h=8 text{hours}$. First, we can substitute the value of $color{#4257b2}P$ and $color{#4257b2}h$ in the formula of the mass $color{#4257b2}M(t)$ and then substitute each value of $color{#4257b2}M$ to calculate the time.

$$
M(t)=Pleft(dfrac{1}{2}right)^frac{t}{h}
$$

$$
M(t)=300left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Now we will calculate the time for each requirement as follows:

(a) For the mass $color{#4257b2}200 text{g}$

Now we know that the mass $color{#4257b2}M(t)=200 text{g}$, so we can substitute this value in the formula to find the value of $color{#4257b2}t$.

$$
M(t)=300left(dfrac{1}{2}right)^{frac{t}{8}}
$$

$$
200=300left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Now we can divide the two sides by $color{#4257b2}300$.

$$
dfrac{200}{300}=dfrac{300left(dfrac{1}{2}right)^{frac{t}{8}}}{300}
$$

$$
dfrac{2}{3}=left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Now we note the two sides have different bases, so we can take $color{#4257b2}log_{frac{1}{2}}$ for each side to simplify the right side.

$$
log_{frac{1}{2}} left(dfrac{2}{3}right)=log_{frac{1}{2}} left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Now we note that the right is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log_{frac{1}{2}} left(dfrac{2}{3}right)=dfrac{t}{8} log_{frac{1}{2}} dfrac{1}{2}
$$

Now we can use the property of logarithm $color{#4257b2}log_{a} a=1$ to replace $color{#4257b2}log_{frac{1}{2}} dfrac{1}{2}$ by $color{#4257b2}1$.

$$
log_{frac{1}{2}} left(dfrac{2}{3}right)=dfrac{t}{8}cdot (1)
$$

$$
log_{frac{1}{2}} left(dfrac{2}{3}right)=dfrac{t}{8}
$$

Now we can multiply the two sides by $color{#4257b2}8$.

$$
8 log_{frac{1}{2}} dfrac{2}{3}=left(dfrac{t}{8}right)cdot (8)
$$

$$
8 log_{frac{1}{2}} dfrac{2}{3}=t
$$

Now we can use the calculator to determine $color{#4257b2}log_{frac{1}{2}} dfrac{2}{3}$ to find the value of $color{#4257b2}t$.

$$
8cdot (0.585)=t
$$

$$
4.68=t
$$

So the time which the substance will take to decay to mass $color{#4257b2}200 text{g}$ is $boxed{ t=4.68 text{hours} }$

(b) For the mass $color{#4257b2}100 text{g}$

Now we know that the mass $color{#4257b2}M(t)=100 text{g}$, so we can substitute this value in the formula to find the value of $color{#4257b2}t$.

$$
M(t)=300left(dfrac{1}{2}right)^{frac{t}{8}}
$$

$$
100=300left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Now we can divide the two sides by $color{#4257b2}300$.

$$
dfrac{100}{300}=dfrac{300left(dfrac{1}{2}right)^{frac{t}{8}}}{300}
$$

$$
dfrac{1}{3}=left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Now we note the two sides have different bases, so we can take $color{#4257b2}log_{frac{1}{2}}$ for each side to simplify the right side.

$$
log_{frac{1}{2}} left(dfrac{1}{3}right)=log_{frac{1}{2}} left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Now we note that the right is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log_{frac{1}{2}} left(dfrac{1}{3}right)=dfrac{t}{8} log_{frac{1}{2}} dfrac{1}{2}
$$

Now we can use the property of logarithm $color{#4257b2}log_{a} a=1$ to replace $color{#4257b2}log_{frac{1}{2}} dfrac{1}{2}$ by $color{#4257b2}1$.

$$
log_{frac{1}{2}} left(dfrac{1}{3}right)=dfrac{t}{8}cdot (1)
$$

$$
log_{frac{1}{2}} left(dfrac{1}{3}right)=dfrac{t}{8}
$$

Now we can multiply the two sides by $color{#4257b2}8$.

$$
8 log_{frac{1}{2}} dfrac{1}{3}=left(dfrac{t}{8}right)cdot (8)
$$

$$
8 log_{frac{1}{2}} dfrac{1}{3}=t
$$

Now we can use the calculator to determine $color{#4257b2}log_{frac{1}{2}} dfrac{1}{3}$ to find the value of $color{#4257b2}t$.

$$
8cdot (1.585)=t
$$

$$
12.68=t
$$

So the time which the substance will take to decay to mass $color{#4257b2}100 text{g}$ is $boxed{ t=12.68 text{hours} }$

(c) For the mass $color{#4257b2}75 text{g}$

Now we know that the mass $color{#4257b2}M(t)=100 text{g}$, so we can substitute this value in the formula to find the value of $color{#4257b2}t$.

$$
M(t)=300left(dfrac{1}{2}right)^{frac{t}{8}}
$$

$$
75=300left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Now we can divide the two sides by $color{#4257b2}300$.

$$
dfrac{75}{300}=dfrac{300left(dfrac{1}{2}right)^{frac{t}{8}}}{300}
$$

$$
dfrac{1}{4}=left(dfrac{1}{2}right)^{frac{t}{8}}
$$

But we know that $color{#4257b2}4=2^{2}$, so we can replace $color{#4257b2}4$ from the left side by $color{#4257b2}2^{2}$ to simplify the left side.

$$
dfrac{1}{2^{2}}=left(dfrac{1}{2}right)^{frac{t}{8}}
$$

$$
dfrac{1^{2}}{2^{2}}=left(dfrac{1}{2}right)^{frac{t}{8}}
$$

$$
left(dfrac{1}{2}right)^{2}=left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Note that we rewrite $color{#4257b2}1$ as $color{#4257b2}1^{2}$ to make the numerator and denominator have the same exponent. Now we note that the two sides have the same base $color{#4257b2}dfrac{1}{2}$, so the exponent of the left side must equal the exponent of the right side.

$$
left(dfrac{1}{2}right)^{2}=left(dfrac{1}{2}right)^{frac{t}{8}}
$$

$$
2=dfrac{t}{8}
$$

Now we can multiply the two sides by $color{#4257b2}8$.

$$
(2)cdot (8)=left(dfrac{t}{8}right)cdot (8)
$$

$$
16=t
$$

So the time which the substance will take to decay to mass $color{#4257b2}75 text{g}$ is $boxed{ t=16 text{hours} }$

(d) For the mass $color{#4257b2}20 text{g}$

Now we know that the mass $color{#4257b2}M(t)=20 text{g}$, so we can substitute this value in the formula to find the value of $color{#4257b2}t$.

$$
M(t)=300left(dfrac{1}{2}right)^{frac{t}{8}}
$$

$$
20=300left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Now we can divide the two sides by $color{#4257b2}300$.

$$
dfrac{20}{300}=dfrac{300left(dfrac{1}{2}right)^{frac{t}{8}}}{300}
$$

$$
dfrac{1}{15}=left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Now we note the two sides have different bases, so we can take $color{#4257b2}log_{frac{1}{2}}$ for each side to simplify the right side.

$$
log_{frac{1}{2}} left(dfrac{1}{15}right)=log_{frac{1}{2}} left(dfrac{1}{2}right)^{frac{t}{8}}
$$

Now we note that the right is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log_{frac{1}{2}} left(dfrac{1}{15}right)=dfrac{t}{8} log_{frac{1}{2}} dfrac{1}{2}
$$

Now we can use the property of logarithm $color{#4257b2}log_{a} a=1$ to replace $color{#4257b2}log_{frac{1}{2}} dfrac{1}{2}$ by $color{#4257b2}1$.

$$
log_{frac{1}{2}} left(dfrac{1}{15}right)=dfrac{t}{8}cdot (1)
$$

$$
log_{frac{1}{2}} left(dfrac{1}{15}right)=dfrac{t}{8}
$$

Now we can multiply the two sides by $color{#4257b2}8$.

$$
8 log_{frac{1}{2}} dfrac{1}{15}=left(dfrac{t}{8}right)cdot (8)
$$

$$
8 log_{frac{1}{2}} dfrac{1}{15}=t
$$

Now we can use the calculator to determine $color{#4257b2}log_{frac{1}{2}} dfrac{1}{15}$ to find the value of $color{#4257b2}t$.

$$
8cdot (3.907)=t
$$

$$
31.25=t
$$

So the time which the substance will take to decay to mass $color{#4257b2}20 text{g}$ is $boxed{ t=31.25 text{hours} }$

Large{$text{$text{color{#c34632}(a) $t=4.68 text{hours}$ (c) $t=16 text{hours}$
\
\
Large{color{#c34632}(b) $t=12.68 text{hours}$ (d) $t=31.25 text{hours}$}$}$

(a) We would like to solve $color{#4257b2}49^{x-1}=7sqrt{7}$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others, so we will simplify the left side and then simplify the right side to make each of them has the same base of the other. We know that $color{#4257b2}49=7^{2}$, so we can replace $color{#4257b2}49$ from the left side by $color{#4257b2}7^{2}$.

$$
49^{x-1}=7sqrt{7}
$$

$$
[7^{2}]^{left(x-1right)}=7sqrt{7}
$$

Now we can use the property of exponents $color{#4257b2}(x^{a})^{b}=x^{ab}$.

$$
7^{(2)cdot (x-1)}=7sqrt{7}
$$

$$
7^{2x-2}=7sqrt{7}
$$

Now we simplified the left side to be with the base $color{#4257b2}7$, so the next step is to simplify the right side to have the same base. We know that $color{#4257b2}sqrt{x}=x^{frac{1}{2}}$, so we can replace $color{#4257b2}sqrt{7}$ from the right side by $color{#4257b2}7^{frac{1}{2}}$.

$$
7^{2x-2}=7sqrt{7}
$$

$$
7^{2x-2}=7cdot (7)^{frac{1}{2}}
$$

Now we can use the property of exponents $color{#4257b2}x^{a}cdot x^{b}=x^{ab}$.

$$
7^{2x-2}=7^{1+frac{1}{2}}
$$

$$
7^{2x-2}=7^{frac{3}{2}}
$$

Now the two sides have the same base $color{#4257b2}7$, so the exponent of the left side will equal the exponent of the right side.

$$
2x-2=dfrac{3}{2}
$$

$$
2x=dfrac{3}{2}+2
$$

$$
2x=dfrac{7}{2}
$$

Now we can divide the two sides by $color{#4257b2}2$ to find the value of $color{#4257b2}x$.

$$
dfrac{2x}{2}=dfrac{dfrac{7}{2}}{2}
$$

$$
x=dfrac{7}{4}
$$

So the solution of the equation is $boxed{ x=dfrac{7}{4} }$

(b) We would like to solve $color{#4257b2}2^{3x-4}=0.25$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others, so we will simplify the right side to make it has the same base of the left side. The first step is to rewrite $color{#4257b2}0.25$ as $color{#4257b2}dfrac{1}{4}$ to make it on the fraction form.

$$
2^{3x-4}=0.25
$$

$$
2^{3x-4}=dfrac{1}{4}
$$

We know that $color{#4257b2}4=2^{2}$, so we can replace $color{#4257b2}4$ from the right side by $color{#4257b2}2^{2}$.

$$
2^{3x-4}=dfrac{1}{2^{2}}
$$

Now we can use the property of exponents $color{#4257b2}dfrac{1}{x^{a}}=x^{-a}$ to replace $color{#4257b2}dfrac{1}{2^{2}}$ by $color{#4257b2}2^{-2}$.

$$
2^{3x-4}=2^{-2}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
3x-4=-2
$$

$$
3x=-2+4=2
$$

Now we can divide the two sides by $color{#4257b2}3$ to find the value of $color{#4257b2}x$.

$$
dfrac{3x}{3}=dfrac{2}{3}
$$

$$
x=dfrac{2}{3}
$$

So the solution of the equation is $boxed{ x=dfrac{2}{3} }$

(c) We would like to solve $color{#4257b2}left(dfrac{1}{4}right)^{x+4}=sqrt{8}$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others, so we will simplify the left side and then simplify the right side to make each of them has the same base of the other. We know that $color{#4257b2}4=2^{2}$, so we can replace $color{#4257b2}4$ from the left side by $color{#4257b2}2^{2}$.

$$
left(dfrac{1}{4}right)^{x+4}=sqrt{8}
$$

$$
left(dfrac{1}{2^{2}}right)^{x+4}=sqrt{8}
$$

Now we can use the property of exponents $color{#4257b2}dfrac{1}{x^{a}}=x^{-a}$ to replace $color{#4257b2}dfrac{1}{2^{2}}$ by $color{#4257b2}2^{-2}$.

$$
left(2^{-2}right)^{x+4}=sqrt{8}
$$

Now we can use the property of exponents $color{#4257b2}(x^{a})^{b}=x^{ab}$ to simplify $color{#4257b2}left(2^{-2}right)^{x+4}$.

$$
2^{(-2)cdot (x+4)}=sqrt{8}
$$

$$
2^{-2x-8}=sqrt{8}
$$

Now we simplified the left side to be with the base $color{#4257b2}2$, so the next step is to simplify the right side to have the same base. We know that $color{#4257b2}8=2^{3}$, so we can replace $color{#4257b2}8$ from the right side by $color{#4257b2}2^{3}$.

$$
2^{-2x-8}=sqrt{2^{3}}
$$

But we know that $color{#4257b2}sqrt{x^{a}}=x^{frac{a}{2}}$, so we can replace $color{#4257b2}sqrt{2^{3}}$ from the right side by $color{#4257b2}2^{frac{3}{2}}$.

$$
2^{-2x-8}=sqrt{2^{3}}
$$

$$
2^{-2x-8}=2^{frac{3}{2}}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
-2x-8=dfrac{3}{2}
$$

$$
-2x=dfrac{3}{2}+8
$$

$$
-2x=dfrac{19}{2}
$$

Now we can divide the two sides by $color{#4257b2}-2$ to find the value of $color{#4257b2}x$.

$$
dfrac{-2x}{-2}=dfrac{dfrac{19}{2}}{-2}
$$

$$
x=-dfrac{19}{4}
$$

So the solution of the equation is $boxed{ x=-dfrac{19}{4} }$

(d) We would like to solve $color{#4257b2}36^{2x+4}=left(sqrt{1296}right)^{x}$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others, so we will simplify the right side to make it has the same base of the left side. First if we used the calculator to determine $color{#4257b2}sqrt{1296}$ we will find that $color{#4257b2}sqrt{1296}=36$, so we can replace $color{#4257b2}sqrt{1296}$ from the right side by $color{#4257b2}36$.

$$
36^{2x+4}=left(sqrt{1296}right)^{x}
$$

$$
36^{2x+4}=left(36right)^{x}
$$

Now the two sides have the same base $color{#4257b2}36$, so the exponent of the left side will equal the exponent of the right side.

$$
2x+4=x
$$

Now we can subtract $color{#4257b2}x+4$ from each side to make $color{#4257b2}x$ in the left side alone.

$$
2x+4-x-4=x-x-4
$$

$$
2xcancel{+4}-xcancel{-4}=cancel{x}cancel{-x}-4
$$

$$
2x-x=-4
$$

$$
x=-4
$$

So the solution of the equation is $boxed{ x=-4 }$

(e) We would like to solve $color{#4257b2}2^{2x+2}+7=71$. First, we can subtract $color{#4257b2}7$ from each side to make.

$$
2^{2x+2}+7=71
$$

$$
2^{2x+2}=71-7
$$

$$
2^{2x+2}=64
$$

Now we need to make the two sides have the same base because in this case the two exponents will equal each others, so we will simplify the right side to make it has the same base of the left side. We know that $color{#4257b2}64=2^{6}$, so we can replace $color{#4257b2}64$ from the right side by $color{#4257b2}2^{6}$.

$$
2^{2x+2}=64
$$

$$
2^{2x+2}=2^{6}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
2x+2=6
$$

$$
2x=6-2=4
$$

Now we can divide the two sides by $color{#4257b2}2$ to find the value of $color{#4257b2}x$.

$$
dfrac{2x}{2}=dfrac{4}{2}
$$

$$
x=2
$$

So the solution of the equation is $boxed{ x=2 }$

(f) We would like to solve $color{#4257b2}9^{2x+1}=81(27^{x})$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others, so we will simplify the left side and then simplify the right side to make each of them has the same base of the other. We know that $color{#4257b2}9=3^{2}$, so we can replace $color{#4257b2}9$ from the left side by $color{#4257b2}3^{2}$.

$$
9^{2x+1}=81(27^{x})
$$

$$
3^{2left(2x+1right)}=81(27^{x})
$$

$$
3^{4x+2}=81(27^{x})
$$

Now we simplified the left side to be with the base $color{#4257b2}3$, so the next step is to simplify the right side to have the same base. We know that $color{#4257b2}27=3^{3}$ and $color{#4257b2}81=3^{4}$, so we can use these facts to simplify the right side.

$$
3^{4x+2}=81(27^{x})
$$

$$
3^{4x+2}=left(3^{4}right)cdot [3^{3}]^{x}
$$

Now we can use the property of exponents $color{#4257b2}(x^{a})^{b}=x^{ab}$.

$$
3^{4x+2}=left(3^{4}right)cdot 3^{3x}
$$

Now we can use the property of exponents $color{#4257b2}x^{a}cdot x^{b}=x^{ab}$.

$$
3^{4x+2}=3^{4+3x}
$$

Now the two sides have the same base $color{#4257b2}3$, so the exponent of the left side will equal the exponent of the right side.

$$
4x+2=4+3x
$$

Now we can subtract $color{#4257b2}3x+2$ from each side to make $color{#4257b2}x$ in the left side alone.

$$
4x+2-3x-2=4+3x-3x-2
$$

$$
4x-3x=4-2
$$

$$
x=2
$$

So the solution of the equation is $boxed{ x=2 }$

Large{$text{$text{color{#c34632}(a) $x=dfrac{7}{4}$ (c) $x=-dfrac{19}{4}$ (e) $2$
\
\
\
Large{color{#c34632}(b) $x=dfrac{2}{3}$ (d) $x=-4$ (f) $2$}$}$

(a) We would like to determine the time it will take for $color{#4257b2}$500$ which is deposited into an account that pays $color{#4257b2}8%$/a compounded annually to be doubled. First, we can use the formula $color{#4257b2}A=Pleft(1+iright)^{n}$, where $color{#4257b2}A$ is the money after the increase, $color{#4257b2}P$ is the fixed money, $color{#4257b2}i$ is the rate of increasing per year and $color{#4257b2}n$ is the number of years. Now from the information of the problem we can substitute $color{#4257b2}P=500, A=1000$ and $color{#4257b2}i=.08$ to find $color{#4257b2}n$.

$$
A=Pleft(1+iright)^{n}
$$

$$
1000=500left(1+.08right)^{n}
$$

$$
1000=500left(1.08right)^{n}
$$

Now we can divide the two sides by $color{#4257b2}500$.

$$
dfrac{1000}{500}=dfrac{500left(1.08right)^{n}}{500}
$$

$$
2=left(1.08right)^{n}
$$

Now we note that the two sides have different bases, so we can take $color{#4257b2}log$ for each side to simplify.

$$
log 2=log (1.08)^{n}
$$

Now we note that the right side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log 2=n log 1.08
$$

Now we can divide the two sides by $color{#4257b2}log 1.08$ to find the value of $color{#4257b2}n$.

$$
dfrac{log 2}{log 1.08}=dfrac{n log 1.08}{log 1.08}
$$

$$
dfrac{log 2}{log 1.08}=n
$$

$$
9.006=n
$$

$$
nsimeq 9
$$

So the time it will take to be doubled is $boxed{ n=9 text{years} }$

(b) We would like to determine the time it will take for a $color{#4257b2}$1000$ investment which is made in a trust fund that pays $color{#4257b2}12%$/a compounded monthly to grow to $color{#4257b2}$5000$. First, we can use the formula $color{#4257b2}A=Pleft(1+iright)^{n}$, where $color{#4257b2}A$ is the money after the increase, $color{#4257b2}P$ is the fixed money of the investment, $color{#4257b2}i$ is the rate of increasing per month and $color{#4257b2}n$ is the number of months. Now from the information of the problem we can substitute $color{#4257b2}P=1000, A=5000$ and $color{#4257b2}i=.12$ to find $color{#4257b2}n$.

$$
A=Pleft(1+iright)^{n}
$$

$$
5000=1000left(1+.12right)^{n}
$$

$$
5000=1000left(1.12right)^{n}
$$

Now we can divide the two sides by $color{#4257b2}1000$.

$$
dfrac{5000}{1000}=dfrac{1000left(1.12right)^{n}}{1000}
$$

$$
5=left(1.12right)^{n}
$$

Now we note that the two sides have different bases, so we can take $color{#4257b2}log$ for each side to simplify.

$$
log 5=log (1.12)^{n}
$$

Now we note that the right side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log 5=n log 1.12
$$

Now we can divide the two sides by $color{#4257b2}log 1.12$ to find the value of $color{#4257b2}n$.

$$
dfrac{log 5}{log 1.12}=dfrac{n log 1.12}{log 1.12}
$$

$$
dfrac{log 5}{log 1.12}=n
$$

$$
14.2=n
$$

$$
nsimeq 14
$$

So the time it will take to grow to $color{#4257b2}$5000$ is $boxed{ n=14 text{months} }$

(c) We would like to determine the time it will take for a $color{#4257b2}$5000$ investment which is made in a savings account that pays $color{#4257b2}10%$/a compounded quarterly to grow to $color{#4257b2}$7500$. First, we can use the formula $color{#4257b2}A=Pleft(1+iright)^{n}$, where $color{#4257b2}A$ is the money after the increase, $color{#4257b2}P$ is the fixed money of the investment, $color{#4257b2}i$ is the rate of increasing per quarter and $color{#4257b2}n$ is the number of quarter of year. Now from the information of the problem we can substitute $color{#4257b2}P=5000, A=7500$ and $color{#4257b2}i=.1$ to find $color{#4257b2}n$.

$$
A=Pleft(1+iright)^{n}
$$

$$
7500=5000left(1+.1right)^{n}
$$

$$
7500=5000left(1.1right)^{n}
$$

Now we can divide the two sides by $color{#4257b2}5000$.

$$
dfrac{7500}{5000}=dfrac{5000left(1.1right)^{n}}{5000}
$$

$$
1.5=left(1.1right)^{n}
$$

Now we note that the two sides have different bases, so we can take $color{#4257b2}log$ for each side to simplify.

$$
log 1.5=log (1.1)^{n}
$$

Now we note that the right side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log 1.5=n log 1.1
$$

Now we can divide the two sides by $color{#4257b2}log 1.1$ to find the value of $color{#4257b2}n$.

$$
dfrac{log 1.5}{log 1.1}=dfrac{n log 1.1}{log 1.1}
$$

$$
dfrac{log 1.5}{log 1.1}=n
$$

$$
4.25=n
$$

Now we can multiply the result by $color{#4257b2}4$ to determine the number of months.

$$
n=(4.25)cdot (4)
$$

$$
n=17
$$

So the time it will take to grow to $color{#4257b2}$7500$ is $boxed{ n=17 text{months} }$

(d) We would like to determine the time it will take for a $color{#4257b2}$500$ if we invested them into an account that pays $color{#4257b2}12%$/a compounded weekly to be triple. First, we can use the formula $color{#4257b2}A=Pleft(1+iright)^{n}$, where $color{#4257b2}A$ is the money after the increase, $color{#4257b2}P$ is the fixed money, $color{#4257b2}i$ is the rate of increasing per week and $color{#4257b2}n$ is the number of weeks. Now from the information of the problem we can substitute $color{#4257b2}P=500, A=1500$ and $color{#4257b2}i=.12$ to find $color{#4257b2}n$.

$$
A=Pleft(1+iright)^{n}
$$

$$
1500=500left(1+.12right)^{n}
$$

$$
1500=500left(1.12right)^{n}
$$

Now we can divide the two sides by $color{#4257b2}500$.

$$
dfrac{1500}{500}=dfrac{500left(1.12right)^{n}}{500}
$$

$$
3=left(1.12right)^{n}
$$

Now we note that the two sides have different bases, so we can take $color{#4257b2}log$ for each side to simplify.

$$
log 3=log (1.12)^{n}
$$

Now we note that the right side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log 3=n log 1.12
$$

Now we can divide the two sides by $color{#4257b2}log 1.12$ to find the value of $color{#4257b2}n$.

$$
dfrac{log 3}{log 1.12}=dfrac{n log 1.12}{log 1.12}
$$

$$
dfrac{log 3}{log 1.12}=n
$$

$$
9.7=n
$$

$$
nsimeq 10
$$

So the time it will take to be doubled is $boxed{ n=10 text{weeks} }$

Large{$text{$text{color{#c34632}(a) $n=9 text{years}$ (c) $n=-17 text{months}$
\
\
Large{color{#c34632}(b) $n=14 text{months}$ (d) $n=10 text{weeks}$}$}$

We would like to determine the time it will take for a culture of $color{#4257b2}20$ bacteria to grow to population of $color{#4257b2}163840$ if we know that that it doubles every $color{#4257b2}15 text{min}$. First, we can use the formula $color{#4257b2}A=Pleft(1+iright)^{n}$, where $color{#4257b2}A$ is the number of bacteria after increasing, $color{#4257b2}P$ is the base number of bacteria, $color{#4257b2}i$ is the rate of increasing and $color{#4257b2}n$ is the number of minutes. Now the first step is to know the rate of increasing $color{#4257b2}i$ to make us calculate the time it will take to increase, so if we substituted $color{#4257b2}n=16 text{min}$, then the number of bacteria will be doubled to $color{#4257b2}40$ and now we can substitute in the formula to find the rate of increasing.

$$
A=Pleft(1+iright)^{n}
$$

$$
40=20left(1+iright)^{15}
$$

Now we can divide the two sides by $color{#4257b2}20$.

$$
dfrac{40}{20}=dfrac{20left(1+iright)^{15}}{20}
$$

$$
2=left(1+iright)^{15}
$$

Now we can take the 15th root for each side to eliminate the exponent $color{#4257b2}15$ from the right side.

$$
sqrt[15]{2}=sqrt[15]{left(1+iright)^{15}}
$$

$$
sqrt[15]{2}=1+i
$$

$$
sqrt[15]{2}-1=i
$$

$$
i=.047
$$

Now we found the rate of increasing $color{#4257b2}i$, so the next step to substitute the values of $color{#4257b2}P=20, A=163840$ and $color{#4257b2}i=.047$ in the formula to find the time which the bacteria will take to be $color{#4257b2}163840$.

$$
A=Pleft(1+iright)^{n}
$$

$$
163840=20left(1+.047right)^{n}
$$

$$
163840=20left(1.047right)^{n}
$$

Now we can divide the two sides by $color{#4257b2}20$.

$$
dfrac{163840}{20}=dfrac{20left(1.047right)^{n}}{20}
$$

$$
8192=left(1.047right)^{n}
$$

Now we note that the two sides have different bases, so we can take $color{#4257b2}log$ for each side to simplify.

$$
log 8192=log (1.047)^{n}
$$

Now we note that the right side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log 8192=n log 1.047
$$

Now we can divide the two sides by $color{#4257b2}log 1.047$ to find the value of $color{#4257b2}n$.

$$
dfrac{log 8192}{log 1.047}=dfrac{n log 1.047}{log 1.047}
$$

$$
dfrac{log 8192}{log 1.047}=n
$$

$$
195=n
$$

So the time it will take to grow to $color{#4257b2}163840$ is $boxed{ n=195 text{min} }$

Large{$text{color{#c34632}$n=195 text{min.}$}$

(a) We would like to solve $color{#4257b2}4^{x+1}+4^{x}=160$. First, we know from the properties of exponents that $color{#4257b2}x^{a+b}=x^{a}cdot x^{b}$, so we can use this property to replace $color{#4257b2}4^{x+1}$ by $color{#4257b2}4^{x}cdot 4^{1}$.

$$
4^{x+1}+4^{x}=160
$$

$$
4^{x}cdot 4^{1}+4^{x}=160
$$

Now w note that the two terms in the left side each of them contains $color{#4257b2}4^{x}$, so we can take it as a common factor.

$$
4^{x}left(4^{1}+1right)=160
$$

$$
4^{x}cdot (4+1)=160
$$

$$
4^{x}cdot (5)=160
$$

Now e can divide the two sides by $color{#4257b2}5$ to make the variable $color{#4257b2}x$ in the left side alone.

$$
dfrac{4^{x}cdot (5)}{5}=dfrac{160}{5}
$$

$$
4^{x}=32
$$

Now we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}4=2^{2}$ and $color{#4257b2}32=2^{5}$, so we can replace $color{#4257b2}4$ and $color{#4257b2}32$ from our equation by $color{#4257b2}2^{2}$ and $color{#4257b2}2^{5}$.

$$
4^{x}=32
$$

$$
left(2^{2}right)^{x}=2^{5}
$$

Now we can use the property of exponents $color{#4257b2}(x^{a})^{b}=x^{ab}$ to simplify the left side.

$$
2^{(2)cdot (x)}=2^{5}
$$

$$
2^{2x}=2^{5}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
2x=5
$$

$$
x=dfrac{5}{2}
$$

So the solution of the equation is $boxed{ x=dfrac{5}{2} }$

(b) We would like to solve $color{#4257b2}2^{x+2}+2^{x}=320$. First, we know from the properties of exponents that $color{#4257b2}x^{a+b}=x^{a}cdot x^{b}$, so we can use this property to replace $color{#4257b2}2^{x+2}$ by $color{#4257b2}2^{x}cdot 2^{2}$.

$$
2^{x+2}+2^{x}=320
$$

$$
2^{x}cdot 2^{2}+2^{x}=320
$$

Now w note that the two terms in the left side each of them contains $color{#4257b2}2^{x}$, so we can take it as a common factor.

$$
2^{x}left(2^{2}+1right)=320
$$

$$
2^{x}cdot (4+1)=320
$$

$$
2^{x}cdot (5)=320
$$

Now we can divide the two sides by $color{#4257b2}5$ to make the variable $color{#4257b2}x$ in the left side alone.

$$
dfrac{2^{x}cdot (5)}{5}=dfrac{320}{5}
$$

$$
2^{x}=64
$$

Now we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}64=2^{6}$, so we can replace $color{#4257b2}64$ from the right side by $color{#4257b2}2^{6}$.

$$
2^{x}=2^{6}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
x=6
$$

So the solution of the equation is $boxed{ x=6 }$

(c) We would like to solve $color{#4257b2}2^{x+2}-2^{x}=96$. First, we know from the properties of exponents that $color{#4257b2}x^{a+b}=x^{a}cdot x^{b}$, so we can use this property to replace $color{#4257b2}2^{x+2}$ by $color{#4257b2}2^{x}cdot 2^{2}$.

$$
2^{x+2}-2^{x}=96
$$

$$
2^{x}cdot 2^{2}-2^{x}=96
$$

Now w note that the two terms in the left side each of them contains $color{#4257b2}2^{x}$, so we can take it as a common factor.

$$
2^{x}left(2^{2}-1right)=96
$$

$$
2^{x}cdot (4-1)=96
$$

$$
2^{x}cdot (3)=96
$$

Now we can divide the two sides by $color{#4257b2}3$ to make the variable $color{#4257b2}x$ in the left side alone.

$$
dfrac{2^{x}cdot (3)}{3}=dfrac{96}{3}
$$

$$
2^{x}=32
$$

Now we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}32=2^{5}$, so we can replace $color{#4257b2}32$ from the right side by $color{#4257b2}2^{5}$.

$$
2^{x}=2^{5}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
x=5
$$

So the solution of the equation is $boxed{ x=5 }$

(d) We would like to solve $color{#4257b2}10^{x+1}-10^{x}=9000$. First, we know from the properties of exponents that $color{#4257b2}x^{a+b}=x^{a}cdot x^{b}$, so we can use this property to replace $color{#4257b2}10^{x+1}$ by $color{#4257b2}10^{x}cdot 10^{1}$.

$$
10^{x+1}-10^{x}=9000
$$

$$
10^{x}cdot 10^{1}-10^{x}=9000
$$

Now w note that the two terms in the left side each of them contains $color{#4257b2}10^{x}$, so we can take it as a common factor.

$$
10^{x}left(10^{1}-1right)=9000
$$

$$
10^{x}cdot (10-1)=9000
$$

$$
10^{x}cdot (9)=9000
$$

Now we can divide the two sides by $color{#4257b2}9$ to make the variable $color{#4257b2}x$ in the left side alone.

$$
dfrac{10^{x}cdot (9)}{9}=dfrac{9000}{9}
$$

$$
10^{x}=1000
$$

Now we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}1000=10^{3}$, so we can replace $color{#4257b2}1000$ from the right side by $color{#4257b2}10^{3}$.

$$
10^{x}=10^{3}
$$

Now the two sides have the same base $color{#4257b2}10$, so the exponent of the left side will equal the exponent of the right side.

$$
x=3
$$

So the solution of the equation is $boxed{ x=3 }$

(e) We would like to solve $color{#4257b2}3^{x+2}+3^{x}=30$. First, we know from the properties of exponents that $color{#4257b2}x^{a+b}=x^{a}cdot x^{b}$, so we can use this property to replace $color{#4257b2}3^{x+2}$ by $color{#4257b2}3^{x}cdot 3^{2}$.

$$
3^{x+2}+3^{x}=30
$$

$$
3^{x}cdot 3^{2}+3^{x}=30
$$

Now w note that the two terms in the left side each of them contains $color{#4257b2}3^{x}$, so we can take it as a common factor.

$$
3^{x}left(3^{2}+1right)=30
$$

$$
3^{x}cdot (9+1)=30
$$

$$
3^{x}cdot (10)=30
$$

Now we can divide the two sides by $color{#4257b2}10$ to make the variable $color{#4257b2}x$ in the left side alone.

$$
dfrac{3^{x}cdot (10)}{10}=dfrac{30}{10}
$$

$$
3^{x}=3
$$

Now the two sides have the same base $color{#4257b2}3$, so the exponent of the left side will equal the exponent of the right side.

$$
3^{x}=3^{1}
$$

$$
x=1
$$

So the solution of the equation is $boxed{ x=1 }$

(f) We would like to solve $color{#4257b2}4^{x+3}-4^{x}=63$. First, we know from the properties of exponents that $color{#4257b2}x^{a+b}=x^{a}cdot x^{b}$, so we can use this property to replace $color{#4257b2}4^{x+3}$ by $color{#4257b2}4^{x}cdot 4^{3}$.

$$
4^{x+3}-4^{x}=63
$$

$$
4^{x}cdot 4^{3}-4^{x}=63
$$

Now w note that the two terms in the left side each of them contains $color{#4257b2}4^{x}$, so we can take it as a common factor.

$$
4^{x}left(4^{3}-1right)=63
$$

$$
4^{x}cdot (64-1)=63
$$

$$
4^{x}cdot (63)=63
$$

Now we can divide the two sides by $color{#4257b2}63$ to make the variable $color{#4257b2}x$ in the left side alone.

$$
dfrac{4^{x}cdot (63)}{63}=dfrac{63}{63}
$$

$$
4^{x}=1
$$

Now we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}1=4^{0}$, so we can replace $color{#4257b2}1$ from the right side by $color{#4257b2}4^{0}$.

$$
4^{x}=1
$$

$$
4^{x}=4^{0}
$$

Now the two sides have the same base $color{#4257b2}4$, so the exponent of the left side will equal the exponent of the right side.

$$
x=0
$$

So the solution of the equation is $boxed{ x=0 }$

Large{$text{$text{color{#c34632}(a) $x=dfrac{5}{2}$ (c) $x=5$ (e) $1$
\
\
\
Large{color{#c34632}(b) $x=6$ (d) $x=3$ (f) $0$}$}$

(a) We would like to choose a strategy to solve the equation
$color{#4257b2}225 (1.05)^{x}=450$. First, we can divide the two sides by $color{#4257b2}225$ to make the variable $color{#4257b2}x$ in the left side alone.

$$
225 (1.05)^{x}=450
$$

$$
dfrac{225 (1.05)^{x}}{225}=dfrac{450}{225}
$$

$$
(1.05)^{x}=2
$$

Now we note that the two sides have different bases, so the next step is to take $color{#4257b2}log$ for each side.

$$
log (1.05)^{x}=log 2
$$

Now we note that the left side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log (1.05)^{x}=log 2
$$

$$
xlog 1.05=log 2
$$

Now the final step is to divide the two sides by $color{#4257b2}log 1.05$ to find the value of $color{#4257b2}x$.

$$
dfrac{x log 1.05}{log 1.05}=dfrac{log 2}{log 1.05}
$$

$$
x=dfrac{log 2}{log 1.05}
$$

So the strategy to solve the equation $color{#4257b2}225 (1.05)^{x}=450$ consists of four steps and they are: 1) divide the two sides by $color{#4257b2}225$, 2) take $color{#4257b2}log$ for each side, 3) use the power law of logarithms and the final step 4) divide the two sides by $color{#4257b2}log 1.05$

(b) We would like to choose a strategy to solve the equation $color{#4257b2}3^{x+2}+3^{x}=270$. First, we can take $color{#4257b2}3^{x}$ as a common factor from the left side.

$$
3^{x+2}+3^{x}=270
$$

$$
3^{x}left(3^{2}+1right)=270
$$

$$
3^{x}cdot (9+1)=270
$$

$$
3^{x}cdot (10)=270
$$

Now we can divide the two sides by $color{#4257b2}10$.

$$
dfrac{3^{x}cdot (10)}{10}=dfrac{270}{10}
$$

$$
3^{x}=27
$$

But we know that $color{#4257b2}27=3^{3}$, so we can replace $color{#4257b2}27$ from the right side $color{#4257b2}3^{3}$ to make the two sides have the same base.

$$
3^{x}=3^{3}
$$

Now the final step is make the exponent of the left side equals the exponent of the right side because the base of the two sides are equal.

So the strategy to solve the equation $color{#4257b2}3^{x+2}+3^{x}=270$ consists of four steps and they are: 1) take $color{#4257b2}3^{x}$ as a common factor from the left side, 2) divide the two sides by $color{#4257b2}10$, 3) replace $color{#4257b2}27$ by $color{#4257b2}3^{3}$ and the final step is 4) the exponent of the left side equals the exponent of the right side because the base of the two sides are equal.

(a) 1) divide the two sides by $color{#c34632}225$ 2) take $color{#c34632}log$ for each side,

3) use the power law of logarithms 4) divide the two sides by $color{#c34632}log 1.05$

(b) 1) take $color{#c34632}3^{x}$ as a common factor 2) divide the two sides by $color{#c34632}10$,

3) replace $color{#c34632}27$ by $color{#c34632}3^{3}$ 4) equaling the exponents of the two sides

(a) We would like to solve $color{#4257b2}5^{t-1}=3.92$. First, we note that the two sides have different bases, so we can take $color{#4257b2}log_{5}$ for each side to simplify the left side.

$$
5^{t-1}=3.92
$$

$$
log_{5} 5^{t-1}=log_{5} 3.92
$$

Now we note that the left side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log_{5} 5^{t-1}=log_{5} 3.92
$$

$$
(t-1) log_{5} 5=log_{5} 3.92
$$

Now we can use the property of logarithm $color{#4257b2}log_{a} a=1$, to replace $color{#4257b2}log_{5} 5$ by $color{#4257b2}1$.

$$
(t-1)cdot (1)=log_{5} 3.92
$$

$$
t-1=log_{5} 3.92
$$

Now we can add $color{#4257b2}1$ to each side to make the variable $color{#4257b2}t$ in the left side alone.

$$
t=log_{5} 3.92 +1
$$

Now we can use the calculator to determine $color{#4257b2}log_{5} 3.92$ to find the value of $color{#4257b2}t$.

$$
t=0.849+1
$$

$$
t=1.849
$$

So the solution of the equation is $boxed{ t=1.849 }$

(b) We would like to solve $color{#4257b2}x=log_{3} 25$. First, we note that the left side contains the variable $color{#4257b2}x$ only, so we can use the calculator to determine $color{#4257b2}log_{3} 25$ to find the value of $color{#4257b2}x$.

$$
x=log_{3} 25
$$

$$
x=2.93
$$

So the solution of the equation is $boxed{ x=2.93 }$

(c) We would like to solve $color{#4257b2}4^{2x}=5^{2x-1}$. First, we can use the property of the exponents $color{#4257b2}x^{a+b}=x^{a}cdot x^{b}$ to simplify the right side.

$$
4^{2x}=5^{2x-1}
$$

$$
4^{2x}=5^{2x}cdot 5^{-1}
$$

$$
4^{2x}=dfrac{5^{2x}}{5}
$$

Now we can multiply the two sides by $color{#4257b2}dfrac{5}{4^{2x}}$ to make the terms which contain the variable $color{#4257b2}x$ in the right side alone.

$$
left(4^{2x}right)cdot left(dfrac{5}{4^{2x}}right)=left(dfrac{5^{2x}}{5}right)cdot left(dfrac{5}{4^{2x}}right)
$$

$$
left(cancel{4^{2x}}right)cdot left(dfrac{5}{cancel{4^{2x}}}right)=left(dfrac{5^{2x}}{cancel{5}}right)cdot left(dfrac{cancel{5}}{4^{2x}}right)
$$

$$
5=dfrac{5^{2x}}{4^{2x}}
$$

Now we can use the property of exponents $color{#4257b2}dfrac{a^{x}}{b^{x}}=left(dfrac{a}{b}right)^{x}$ to simplify the right side.

$$
5=left(dfrac{5}{4}right)^{2x}
$$

Now we note that the two sides have different bases, so we can take $color{#4257b2}log_{5}$ to each side.

$$
log_{5} 5=log_{5} left(dfrac{5}{4}right)^{2x}
$$

Now we note that the right side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log_{5} 5=(2x)cdot log_{5} dfrac{5}{4}
$$

Now we can use the property of logarithm $color{#4257b2}log_{a} a=1$, to replace $color{#4257b2}log_{5} 5$ by $color{#4257b2}1$.

$$
1=(2x)cdot log_{5} 1.25
$$

Now we can divide the two sides by $color{#4257b2}2log_{5} 1.25$ to each side to make the variable $color{#4257b2}x$ in the right side alone.

$$
dfrac{1}{2log_{5} 1.25}=x
$$

Now we can use the calculator to determine $color{#4257b2}log_{5} 1.25$ to find the value of $color{#4257b2}x$.

$$
dfrac{1}{2cdot (0.1387)}=x
$$

$$
3.606=x
$$

So the solution of the equation is $boxed{ x=3.606 }$

(d) We would like to solve $color{#4257b2}x=log_{2} 53.2$. First, we note that the left side contains the variable $color{#4257b2}x$ only, so we can use the calculator to determine $color{#4257b2}log_{2} 53.2$ to find the value of $color{#4257b2}x$.

$$
x=log_{2} 53.2
$$

$$
x=5.733
$$

So the solution of the equation is $boxed{ x=5.733 }$

Large{$text{$text{color{#c34632}(a) $t=1.849$ (c) $x=3.606$
\
\
Large{color{#c34632}(b) $x=2.93$ (d) $x=5.733$}$}$

We would like to solve the equation $color{#4257b2}3^{2x}-5 (3^{x})=-6$. First, we can add $color{#4257b2}6$ to each side to make the right side equals zero.

$$
3^{2x}-5 (3^{x})=-6
$$

$$
3^{2x}-5 (3^{x})+6=0
$$

Now we can let $color{#4257b2}y=3^{x}$, then $color{#4257b2}y^{2}=(3^{x})^{2}=3^{2x}$ where we know from the properties of exponents that $color{#4257b2}(x^{a})^{b}=x^{ab}$. Now we can substitute our supposition in the equation to simplify.

$$
3^{2x}-5 (3^{x})+6=0
$$

$$
y^{2}-5 y+6=0
$$

Note that we made this supposition to convert our equation to a quadratic equation which we solve it easily. Now we have a quadratic equation, so we can factor to find the values of $color{#4257b2}y$.

$$
y^{2}-5 y+6=0
$$

$$
(y-2)(y-3)=0
$$

Now we can use the zero-factor property.

$$
y-2=0 text{or} y-3=0
$$

$$
y=2 text{or} y=3
$$

Now we can do back substitution where $color{#4257b2}y=3^{x}$ to find the values of $color{#4257b2}x$.

$$
3^{x}=2 text{or} 3^{x}=3
$$

Now we have two cases, so we can solve each of them to find the values of $color{#4257b2}x$ as follows:

For $color{#4257b2}3^{x}=2$

We note that we have two different bases, so we can take $color{#4257b2}log_{3}$ for each side.

$$
log_{3} 3^{x}=log_{3} 2
$$

Now we note that the left side is a power of logarithms, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log_{3} 3^{x}=log_{3} 2
$$

$$
x log_{3} 3=log_{3} 2
$$

Now we can use the property of logarithms $color{#4257b2}log_{a} a=1$ to replace $color{#4257b2}log_{3} 3$ by $color{#4257b2}1$.

$$
xcdot (1)=log_{3} 2
$$

$$
x=log_{3} 2
$$

Now we can use the calculator to determine $color{#4257b2}log_{3} 2$ to find the value of $color{#4257b2}x$.

$$
x=0.631
$$

So the solution of the first case is $boxed{ x=0.631 }$

For $color{#4257b2}3^{x}=3$

We note that the two sides have the same base $color{#4257b2}3$, so the exponent of the left side must equal the exponent of the right side.

$$
3^{x}=3
$$

$$
3^{x}=3^{1}
$$

$$
x=1
$$

So the solution of the second case is $boxed{ x=1 }$

Now we found the solution of each case, so the solutions of the equation are $color{#4257b2}x=0.631 {color{Black}text{or}} x=1$

Large{$text{color{#c34632}$x=0.631 {color{Black}text{or}} x=1$}$

We would like to show that $color{#4257b2}y=dfrac{log x}{log a}$ if we know that $color{#4257b2}log_{a} x=y$. First, we can convert the equation from the logarithmic form to the exponential form where if $color{#4257b2}a=log_{b} x$, then $color{#4257b2}x=b^{a}$.

$$
log_{a} x=y
$$

$$
a^{y}=x
$$

Now our equation became on the exponential form, so the next step is to take $color{#4257b2}log$ for each side.

$$
a^{y}=x
$$

$$
log a^{y}=log x
$$

Now we note that the left side is a power of logarithms, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log a^{y}=log x
$$

$$
y log a=log x
$$

Now we can divide the two sides by $color{#4257b2}log a$.

$$
dfrac{y log a}{log a}=dfrac{log x}{log a}
$$

$$
y=dfrac{log x}{log a}
$$

So we proved that if $color{#4257b2}log_{a} x=y$, then $boxed{ y=dfrac{log x}{log a} }$

Now we can use this relationship to simplify $color{#4257b2}y=log_{5} x$ and then explain who to graph it. First, we note that the equation $color{#4257b2}y=log_{5} x$ is on the same form of the equation $color{#4257b2}y=log_{a} x$ where $color{#4257b2}a=5$, so we can use the relationship which we proved as follows:

$$
y=log_{5} x
$$

$$
y=dfrac{log x}{log 5}
$$

Now we can use the calculator to determine the value of $color{#4257b2}log 5$.

$$
y=dfrac{log x}{0.7}
$$

$$
y=1.431 log x
$$

Now we note that our function is a logarithm function which multiplied by $color{#4257b2}1.431$, so we can graph it using the graphing calculator.

Large{$text{color{#c34632}$y=dfrac{log x}{log a}$}$

(a) We would like to solve $color{#4257b2}2^{x^{2}}=32 (2^{4x})$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}32=2^{5}$, so we can replace $color{#4257b2}32$ from the right side by $color{#4257b2}2^{5}$.

$$
2^{x^{2}}=32 (2^{4x})
$$

$$
2^{x^{2}}=(2^{5})cdot (2^{4x})
$$

Now we can use the property of exponents $color{#4257b2}x^{a}cdot x^{b}=x^{a+b}$ to simplify the right side.

$$
2^{x^{2}}=2^{5+4x}
$$

Now the two sides have the same base $color{#4257b2}2$, so the exponent of the left side will equal the exponent of the right side.

$$
x^{2}=5+4x
$$

Now we can subtract $color{#4257b2}5+4x$ from each side to make the right side equals zero.

$$
x^{2}-(5+4x)=0
$$

$$
x^{2}-4x-5=0
$$

Now we have a quadratic equation, so we can factor to find the values of $color{#4257b2}x$.

$$
x^{2}-4x-5=0
$$

$$
(x-5)(x+1)=0
$$

Now we can use the zero-factor property.

$$
x-5=0 text{or} x+1=0
$$

$$
x=5 text{or} x=-1
$$

So the solutions of the equation are $boxed{ x=5 } text{or} boxed{ x=-1 }$

(b) We would like to solve $color{#4257b2}3^{x^{2}+20}=left(dfrac{1}{27}right)^{3x}$. First, we need to make the two sides have the same base because in this case the two exponents will equal each others. We know that $color{#4257b2}27=3^{3}$, so we can replace $color{#4257b2}27$ from the right side by $color{#4257b2}3^{3}$.

$$
3^{x^{2}+20}=left(dfrac{1}{27}right)^{3x}
$$

$$
3^{x^{2}+20}=left(dfrac{1}{3^{3}}right)^{3x}
$$

Now we can use the property of exponent $color{#4257b2}dfrac{1}{x^{a}}=x^{-a}$ to replace $color{#4257b2}dfrac{1}{3^{3}}$ from the right side by $color{#4257b2}3^{-3}$.

$$
3^{x^{2}+20}=left(3^{-3}right)^{3x}
$$

Now we can use the property of exponents $color{#4257b2}(x^{a})^{b}=x^{ab}$ to simplify the right side.

$$
3^{x^{2}+20}=3^{(-3)cdot (3x)}
$$

$$
3^{x^{2}+20}=3^{-9x}
$$

Now the two sides have the same base $color{#4257b2}3$, so the exponent of the left side will equal the exponent of the right side.

$$
x^{2}+20=-9x
$$

Now we can add $color{#4257b2}9x$ to each side to make the right side equals zero.

$$
x^{2}+20+9x=-9x+9x
$$

$$
x^{2}+9x+20=0
$$

Now we have a quadratic equation, so we can factor to find the values of $color{#4257b2}x$.

$$
x^{2}+9x+20=0
$$

$$
(x+4)(x+5)=0
$$

Now we can use the zero-factor property.

$$
x+4=0 text{or} x+5=0
$$

$$
x=-4 text{or} x=-5
$$

So the solutions of the equation are $boxed{ x=-4 } text{or} boxed{ x=-5 }$

(c) We would like to solve $color{#4257b2}(2)cdot (3^{x})=(7)cdot (5^{x})$. First, we can divide the two sides by $color{#4257b2}2 (5^{x})$ to make the terms which contain the variable $color{#4257b2}x$ in the left side alone.

$$
(2)cdot (3^{x})=(7)cdot (5^{x})
$$

$$
dfrac{(2)cdot (3^{x})}{2 (5^{x})}=dfrac{(7)cdot (5^{x})}{2 (5^{x})}
$$

$$
dfrac{cancel{(2)}cdot (3^{x})}{cancel{2} (5^{x})}=dfrac{(7)cdot cancel{(5^{x})}}{2 cancel{(5^{x})}}
$$

$$
dfrac{3^{x}}{5^{x}}=dfrac{7}{2}
$$

$$
dfrac{3^{x}}{5^{x}}=3.5
$$

Now we can use the property of exponents $color{#4257b2}dfrac{a^{x}}{b^{x}}=left(dfrac{a}{b}right)^{x}$ to simplify the left side.

$$
left(dfrac{3}{5}right)^{x}=3.5
$$

$$
(0.6)^{x}=3.5
$$

Now we note that the two sides have different bases, so we can take $color{#4257b2}log$ to each side.

$$
log (0.6)^{x}=log 3.5
$$

Now we note that the right side is a power of logarithm, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
log (0.6)^{x}=log 3.5
$$

$$
x log 0.6=log 3.5
$$

Now we can divide the two sides by $color{#4257b2}log 0.6$ to find the value of $color{#4257b2}x$.

$$
dfrac{x log 0.6}{log 0.6}=dfrac{log 3.5}{log 0.6}
$$

$$
x=dfrac{log 3.5}{log 0.6}
$$

Now we can use the calculator to determine $color{#4257b2}log 3.5$ and $color{#4257b2}log 0.6$ to find the value of $color{#4257b2}x$.

$$
x=dfrac{0.544}{-0.222}
$$

$$
x=-2.452
$$

So the solution of the equation is $boxed{ x=-2.452 }$

Large{$text{color{#c34632}(a) $x=5 {color{Black}text{or}} x=-1$
\
\
\
(b) $x=-4 {color{Black}text{or}} x=-5$
\
\
\
(c) $x=-2.452$}$

We would like to show that if $color{#4257b2}log_{a} 2=log_{b} 8$, then $color{#4257b2}a^{3}=b$. First, we can suppose that $color{#4257b2}log_{a} 2=log_{b} 8=x$ and then split this supposition to two equations $color{#4257b2}log_{a} 2=x$ and $color{#4257b2}log_{b} 8=x$.

$$
log_{a} 2=log_{b} 8=x
$$

$$
log_{a} 2=x text{and} log_{b} 8=x
$$

Now we can simplify each equation as follows:

For $color{#4257b2}log_{a} 2=x$

We can convert this equation from the logarithmic form to the exponential form where if $color{#4257b2}log_{a} x=b$, then $color{#4257b2}x=a^{b}$.

$$
log_{a} 2=x
$$

$$
2=a^{x} (1)
$$

Now we can simplify the second equation.

For $color{#4257b2}log_{b} 8=x$

We can also convert this equation from the logarithmic form to the exponential form where if $color{#4257b2}log_{a} x=b$, then $color{#4257b2}x=a^{b}$.

$$
log_{b} 8=x
$$

$$
8=b^{x}
$$

But we know that $color{#4257b2}8=2^{3}$, so we can replace $color{#4257b2}8$ from the left side by $color{#4257b2}2^{3}$

$$
2^{3}=b^{x}
$$

Now we can substitute the equation (1) in the last equation.

$$
color{#4257b2}2^{3}=b^{x} {color{Black}text{and}} 2=a^{x}
$$

$$
(a^{x})^{3}=b^{x}
$$

But we know from the properties of exponential that $color{#4257b2}(x^{a})^{b}=(x^{b})^{a}$, so we can use this property to simplify the left side.

$$
(a^{x})^{3}=b^{x}
$$

$$
(a^{3})^{x}=b^{x}
$$

Now the two sides of the equation has the same exponential $color{#4257b2}x$, so the base of the left side must equal the base of the right side.

$$
(a^{3})^{x}=(b)^{x}
$$

$$
a^{3}=b
$$

So we proved that if $color{#4257b2}log_{a} 2=log_{b} 8$, then $boxed{ a^{3}=b }$

Large{$text{color{#c34632}$a^{3}=b$}$

We would like to find the point of intersection between the graphs of
$color{#4257b2}y=3(5^{2x})$ and $color{#4257b2}y=6(4^{3x})$. First, to find the point of intersection between the two graphs we can make the first function equal the second function to find the value of $color{#4257b2}x$ at this point.

$$
because y=3(5^{2x}) text{and} y=6(4^{3x})
$$

$$
3(5^{2x})=6 (4^{3x})
$$

Now we can divide the two sides by $color{#4257b2}3$.

$$
dfrac{3(5^{2x})}{3}=dfrac{6 (4^{3x})}{3}
$$

$$
5^{2x}=2 (4^{3x})
$$

But we know that $color{#4257b2}4=2^{x}$, so we can replace $color{#4257b2}4$ from the right side by $color{#4257b2}2^{2}$.

$$
5^{2x}=2 (2^{2})^{3x}
$$

Now we can use the property of exponents $color{#4257b2}(x^{a})^{b}$ to simplify the right side.

$$
5^{2x}=2 [2^{(2)cdot (3x)}]
$$

$$
5^{2x}=2 (2^{6x})
$$

Now we can use the property of exponents $color{#4257b2}x^{a}cdot x^{b}=x^{a+b}$ to simplify the right side.

$$
5^{2x}=2^{1+6x}
$$

Now we note that the two sides have different bases, so we can take $color{#4257b2}log$ fro each side.

$$
log 5^{2x}=log 2^{1+6x}
$$

Now we note that the two sides is a power of logarithms, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
(2x) log 5=(1+6x) log 2
$$

$$
2x log 5=log 2+6x log 2
$$

Now we can subtract $color{#4257b2}6x log 2$ from each side.

$$
2x log 5-6x log 2=log 2
$$

Now we note that the two terms in the left side each of them contains $color{#4257b2}x$, so we can take it as a common factor.

$$
xleft(2log 5-6log 2right)=log 2
$$

Now we can divide the two sides by $color{#4257b2}2log 5-6log 2$ to find the value of $color{#4257b2}x$.

$$
dfrac{xcancel{left(2log 5-6log 2right)}}{cancel{2log 5-6log 2}}=dfrac{log 2}{2log 5-6log 2}
$$

$$
x=dfrac{log 2}{2log 5-6log 2}
$$

Now we can use the calculator to determine $color{#4257b2}log 2$ and $color{#4257b2}log 5$ to find the value of $color{#4257b2}x$.

$$
x=dfrac{0.301}{2(0.699)-6(0.301)}
$$

$$
x=-0.737
$$

Now we found the value of $color{#4257b2}x$ at the point of intersection, so the next step is to substitute this value in one of the functions of $color{#4257b2}y$ to find the value of $color{#4257b2}y$ at this intersection point.

$$
y=3(5^{2x})
$$

$$
y=3 [5^{(2)cdot (-0.737)}]
$$

$$
y=3 (5^{-1.474})
$$

$$
y=(3)cdot (0.0933)
$$

$$
y=0.28
$$

Now we found the value of $color{#4257b2}x$ and $color{#4257b2}y$ for the point of intersection between the two graphs, so this point is $color{#4257b2}(-0.737, 0.28)$

$$
color{#c34632}(-0.737, 0.28)
$$

(a) We would like to solve $color{#4257b2}6^{3x}=4^{2x-3}$. First, we note that the two sides have different bases, so we can take $color{#4257b2}log$ fro each side.

$$
6^{3x}=4^{2x-3}
$$

$$
log 6^{3x}=log 4^{2x-3}
$$

Now we note that the two sides is a power of logarithms, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
(3x) log 6=(2x-3) log 4
$$

$$
3x log 6=2x log 4-3 log 4
$$

Now we can subtract $color{#4257b2}2x log 4$ from each side.

$$
3x log 6-2x log 4=-3 log 4
$$

Now we note that the two terms in the left side each of them contains $color{#4257b2}x$, so we can take it as a common factor.

$$
xleft(3log 6-2log 4right)=-3log 4
$$

Now we can divide the two sides by $color{#4257b2}3log 6-2log 4$ to find the value of $color{#4257b2}x$.

$$
dfrac{xcancel{left(3log 6-2log 4right)}}{cancel{3log 6-2log 4}}=dfrac{-3log 4}{3log 6-2log 4}
$$

$$
x=dfrac{-3log 4}{3log 6-2log 4}
$$

Now we can use the calculator to determine $color{#4257b2}log 6$ and $color{#4257b2}log 4$ to find the value of $color{#4257b2}x$.

$$
x=dfrac{-3 (0.602)}{3(0.778)-2(0.602)}
$$

$$
x=-1.6
$$

So the solution of the equation is $boxed{ x=-1.6 }$

(b) We would like to solve $color{#4257b2}(1.2)^{x}=(2.8)^{x+4}$. First, we note that the two sides have different bases, so we can take $color{#4257b2}log$ fro each side.

$$
(1.2)^{x}=(2.8)^{x+4}
$$

$$
log (1.2)^{x}=log (2.8)^{x+4}
$$

Now we note that the two sides is a power of logarithms, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
x log 1.2=(x+4) log 2.8
$$

$$
x log 1.2=x log 2.8+4 log 2.8
$$

Now we can subtract $color{#4257b2}x log 2.8$ from each side.

$$
x log 1.2-x log 2.8=4 log 2.8
$$

Now we note that the two terms in the left side each of them contains $color{#4257b2}x$, so we can take it as a common factor.

$$
xleft(log 1.2-log 2.8right)=4log 2.8
$$

Now we can divide the two sides by $color{#4257b2}log 1.2-log 2.8$ to find the value of $color{#4257b2}x$.

$$
dfrac{xcancel{left(log 1.2-log 2.8right)}}{cancel{log 1.2-log 2.8}}=dfrac{4log 2.8}{log 1.2-log 2.8}
$$

$$
x=dfrac{4log 2.8}{log 1.2-log 2.8}
$$

Now we can use the calculator to determine $color{#4257b2}log 2.8$ and $color{#4257b2}log 1.2$ to find the value of $color{#4257b2}x$.

$$
x=dfrac{4 (0.447)}{(0.079)-(0.447)}
$$

$$
x=-4.86
$$

So the solution of the equation is $boxed{ x=-4.86 }$

(c) We would like to solve $color{#4257b2}3 (2^{x})=4^{x+1}$. First, we know from the properties of exponents that $color{#4257b2}x^{a+b}=x^{a}cdot x^{b}$, so we can use this property to simplify the right side as follows:

$$
3 (2^{x})=4^{x+1}
$$

$$
3 (2^{x})=4^{x}cdot 4^{1}
$$

$$
3 (2^{x})=4 (4^{x})
$$

But we know that $color{#4257b2}4=2^{2}$, so we can use this fact in the right side.

$$
3 (2^{x})=4 (2^{2})^{x}
$$

But we know from the properties of exponents that $color{#4257b2}(x^{a})^{b}=x^{ab}$, so we can simplify the right side using this property.

$$
3 (2^{x})=4 [2^{(2)cdot (x)}]
$$

$$
3 (2^{x})=4 (2^{2x})
$$

Now we can subtract $color{#4257b2}3 (2^{x})$ from each side to make the left side equals zero.

$$
0=4 (2^{2x})-3 (2^{x})
$$

Now we can take $color{#4257b2}2^{x}$ as a common factor from the right side.

$$
0=2^{x}left[4 (2^{x})-3right]
$$

Now we can use the zero-factor property.

$$
2^{x}=0 text{or} 4 (2^{x})-3=0
$$

But we know that $color{#4257b2}a^{x} ne 0$, so the solution $color{#4257b2}2^{x}=0$ is refused.

$$
4 (2^{x})-3=0
$$

Now we can add $color{#4257b2}3$ to each side.

$$
4 (2^{x})=3
$$

Now we can divide the two sides by $color{#4257b2}4$.

$$
2^{x}=dfrac{3}{4}=0.75
$$

Now we note that the two sides have different bases, so we can take $color{#4257b2}log$ fro each side.

$$
log 2^{x}=log 0.75
$$

Now we note that the left side is a power of logarithms, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
x log 2=log 0.75
$$

Now we can divide the two sides by $color{#4257b2}log 2$ to find the value of $color{#4257b2}x$.

$$
x=dfrac{log 0.75}{log 2}
$$

Now we can use the calculator to determine $color{#4257b2}log 0.75$ and $color{#4257b2}log 2$ to find the value of $color{#4257b2}x$.

$$
x=dfrac{-0.125}{0.301}
$$

$$
x=-0.42
$$

So the solution of the equation is $boxed{ x=-0.42 }$

Large{$text{color{#c34632}(a) $x=-1.6$
\
\
(b) $x=-4.86$
\
\
(c) $x=-0.42$}$

We would like to solve $color{#4257b2}(2^{x})^{x}=10$. First, we know from the properties of exponents that $color{#4257b2}(x^{a})^{b}=x^{ab}$, so we can use this property to simplify the left side.

$$
(2^{x})^{x}=10
$$

$$
2^{(x)cdot (x)}=10
$$

$$
2^{x^{2}}=10
$$

Now we note that the two sides have different bases, so we can take $color{#4257b2}log$ fro each side.

$$
log 2^{x^{2}}=log 10
$$

But we know that $color{#4257b2}log 10=1$, so we can substitute this values in the right side.

$$
log 2^{x^{2}}=1
$$

Now we note that the left side is a power of logarithms, so we can use the power law of logarithms where $color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
x^{2} log 2=1
$$

Now we can divide the two sides by $color{#4257b2}log 2$ to find the value of $color{#4257b2}x$.

$$
x^{2}=dfrac{1}{log 2}
$$

Now we can use the calculator to determine $color{#4257b2}log 2$ to find the value of $color{#4257b2}x^{2}$.

$$
x^{2}=dfrac{1}{0.301}
$$

$$
x^{2}=3.322
$$

Now we can take the square root for each side to find the value of $color{#4257b2}x$.

$$
x=pm sqrt{3.322}
$$

$$
x=pm 1.82
$$

So the solutions of the equation are $boxed{ x=-1.82 } text{or} boxed{ x=1.82 }$

Large{$text{color{#c34632}$x=-1.82 {color{Black}text{or}} x=1.82$}$