Write the inverse equation in the logarithmic and exponential form for the following terms.

$$
color{#4257b2}text{(a)} y=4^x
$$

According the standard form for the inverse function as follows:

$$
because y=a^x therefore x=a^y text{equivalent to} y=log_a(x)
$$

Inverse equation in the logarithmic form $y=log_4(x)$

Inverse equation in the exponential form $x=4^y$

$$
color{#4257b2}text{(b)} y=log_6(x)
$$

According the standard form for the inverse function as follows:

$$
because y=a^x therefore x=a^y text{equivalent to} y=log_a(x)
$$

Inverse equation in the logarithmic form $x=log_6(y)$

Inverse equation in the exponential form $y=6^x$

$$
text{color{#c34632}(a) $y=log_4(x)$ $x=4^y$
\ \
(b) $x=log_6(y)$ $y=6^x$}
$$

Describe the transformation for the following function.

$$
color{#4257b2}text{(a)} g(x)=log[2(x-4)]+3
$$

$$
k=2 d=4 c=3
$$

$$
color{#4257b2}text{(b)} g(x)=-dfrac{1}{2} log(x+5)-1
$$

$$
a=-dfrac{1}{2} d=-5 c=-1
$$

$$
text{color{#c34632}(a) $k=2 d=4 c=3$
\ \
(b) $a=-dfrac{1}{2} d=-5 c=-1$}
$$

(a) We would like to evaluate $color{#4257b2}log_{3} dfrac{1}{9}$. First, we can let the solution of this expression equals $color{#4257b2}x$ and then try to solve it.

$$
log_{3} dfrac{1}{9}=x
$$

Now we note that we have an equation on the logarithmic form, so we can convert it to the exponential form where we know that if $color{#4257b2}log_{a} x=b$, then $color{#4257b2}a^{b}=x$.

$$
3^{x}=dfrac{1}{9}
$$

But we know that $color{#4257b2}9=3^{2}$, so we can replace $color{#4257b2}9$ from the right side by $color{#4257b2}3^{2}$.

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

Now we can use the property of exponents $color{#4257b2}dfrac{1}{x^{a}}=x^{-a}$ to replace $color{#4257b2}dfrac{1}{3^{2}}$ from the right side by $color{#4257b2}3^{-2}$.

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

Now 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 where we know that if $color{#4257b2}x^{a}=x^{b}$, then $color{#4257b2}a=b$.

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

$$
x=-2
$$

So we find that $color{#4257b2}x=-2$, so the value of $color{#4257b2}log_{3} dfrac{1}{9}$ is $boxed{ -2 }$

(b) We would like to evaluate $color{#4257b2}log_{5} 100-log_{5} 4$. First, we note that our expression is a difference of logarithms, so we can use the quotient law of logarithms where $color{#4257b2}log_{a} x-log_{a} y=log_{a} left(dfrac{x}{y}right)$.

$$
begin{align*}
log_{5} 100-log_{5} 4&=log_{5} left(dfrac{100}{4}right)
\ \
&=log_{5} 25
\ \
&=log_{5} 5^{2}
end{align*}
$$

Now we note that our expression is simplified to $color{#4257b2}log_{5} 5^{2}$ which is a power of logarithms, so we can use the power law of logarithms where
$color{#4257b2}log_{m} x^{r}=r log_{m} x$.

$$
begin{align*}
log_{5} 100-log_{5} 4&=log_{5} 5^{2}
\ \
&=2log_{5} 5
end{align*}
$$

But we know from the properties of logarithms that $color{#4257b2}log_{a} a=1$, so we can replace $color{#4257b2}log_{5} 5$ by $color{#4257b2}1$.

$$
begin{align*}
log_{5} 100-log_{5} 4&=2log_{5} 5
\ \
&=2cdot (1)
\ \
&=2
end{align*}
$$

So the value of $color{#4257b2}log_{5} 100-log_{5} 4$ is $boxed{ 2 }$

Large{$text{color{#c34632}(a) $log_{3} dfrac{1}{9}=-2$ (b) $log_{5} 100-log_{5} 4=2$}$

(a) We would like to evaluate $color{#4257b2}log 15+log 40-log 6$. First, we note that the first two terms of our expression is a sum of logarithms, so we can use the product law of logarithms where $color{#4257b2}log_{a} x+log_{a} y=log_{a} xy$.

$$
begin{align*}
log 15+log 40-log 6&=log (15)cdot (40)-log 6
\ \
&=log 600-log 6
end{align*}
$$

Now we note that our expression is a difference of logarithms, so we can use the quotient law of logarithms where $color{#4257b2}log_{a} x-log_{a} y=log_{a} left(dfrac{x}{y}right)$.

$$
begin{align*}
log 15+log 40-log 6&=log 600-log 6
\ \
&=log left(dfrac{600}{6}right)
\ \
&=log 100
\ \
&=log 10^{2}
end{align*}
$$

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

$$
begin{align*}
log 15+log 40-log 6&=log 10^{2}
\ \
&=2log 10
\ \
&=2cdot (1)
\ \
&=2
end{align*}
$$

Note that we used the property of logarithms $color{#4257b2}log 10=1$. So the value of $color{#4257b2}log 15+log 40-log 6$ is $boxed{ 2 }$

(b) We would like to evaluate $color{#4257b2}log_{7} 343+2log_{7} 49$. First, we know that $color{#4257b2}49=7^{2}$ and $color{#4257b2}343=7^{3}$, so we can replace $color{#4257b2}49$ and $color{#4257b2}343$ from our expression by $color{#4257b2}7^{2}$ and $color{#4257b2}7^{3}$.

$$
log_{7} 343+2log_{7} 49=log_{7} 7^{3}+2log_{7} 7^{2}
$$

Now we note that the second term of our expression is a single logarithms which is multiplied by a number, so we can use the power law of logarithms where $color{#4257b2}r log_{m} x=log_{m} x^{r}$.

$$
begin{align*}
log_{7} 343+2log_{7} 49&=log_{7} 7^{3}+2log_{7} 7^{2}
\ \
&=log_{7} 7^{3}+log_{7} left(7^{2}right)^{2}
\ \
&=log_{7} 7^{3}+log_{7} 7^{(2)cdot (2)}
\ \
&=log_{7} 7^{3}+log_{7} 7^{4}
end{align*}
$$

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

$$
begin{align*}
log_{7} 343+2log_{7} 49&=log_{7} 7^{3}+log_{7} 7^{4}
\ \
&=3log_{7} 7+4log_{7} 7
\ \
&=3cdot (1)+4cdot (1)
\ \
&=3+4
\ \
&=7
end{align*}
$$

Note that we used the property of logarithms $color{#4257b2}log_{a} a=1$ to replace $color{#4257b2}log_{7} 7$ by $color{#4257b2}1$. So the value of $color{#4257b2}log_{7} 343+2log_{7} 49$ is $boxed{ 7 }$

$$
text{color{#c34632}(a) $log 15+log 40-log 6=2$ (b) $log_{7} 343+2log_{7} 49=7$}
$$

We would like to write $color{#4257b2}log_{4} x^{2}+3log_{4} ydfrac{1}{3}-log_{4} x$. First, we note that the second term of our expression is a single logarithm which is multiplied by a number, so we can use the power law of logarithms where $color{#4257b2}r log_{m} x=log_{m} x^{r}$.

$$
begin{align*}
log_{4} x^{2}+3log_{4} ydfrac{1}{3}-log_{4} x&=log_{4} x^{2}+log_{4} left(dfrac{y}{3}right)^{3}-log_{4} x
\ \
&=log_{4} x^{2}+log_{4} dfrac{y^{3}}{27}-log_{4} x
end{align*}
$$

Now we note that the first two terms of our expression is a sum of logarithms, so we can use the product law of logarithms where $color{#4257b2}log_{a} x+log_{a} y=log_{a} xy$.

$$
begin{align*}
log_{4} x^{2}+3log_{4} ydfrac{1}{3}-log_{4} x&=log_{4} x^{2}+log_{4} dfrac{y^{3}}{27}-log_{4} x
\ \
&=log_{4} left(x^{2}right)cdot left(dfrac{y^{3}}{27}right)-log_{4} x
\ \
&=log_{4} dfrac{x^{2} y^{3}}{27}-log_{4} x
end{align*}
$$

Now we note that our expression is a difference of logarithms, so we can use the quotient law of logarithms where $color{#4257b2}log_{a} x-log_{a} y=log_{a} left(dfrac{x}{y}right)$.

$$
begin{align*}
log_{4} x^{2}+3log_{4} ydfrac{1}{3}-log_{4} x&=log_{4} dfrac{x^{2} y^{3}}{27}-log_{4} x
\ \
&=log_{4} dfrac{dfrac{x^{2}y^{3}}{27}}{x}
\ \
&=log_{4} dfrac{xy^{3}}{27}
end{align*}
$$

So the expression $color{#4257b2}log_{4} x^{2}+3log_{4} ydfrac{1}{3}-log_{4} x$ can be written as the single logarithm $boxed{ log_{4} dfrac{xy^{3}}{27} }$

Large{$text{color{#c34632}$log_{4} dfrac{xy^{3}}{27}$}$

We would like to solve $color{#4257b2}5^{x+2}=6^{x+1}$. First, we note that the two sides have different bases, so we can take $color{#4257b2}log$ for each side to simplify the left side.

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

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

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

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

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

$$
xlog 5+2log 5=xlog 6+log 6
$$

Now we can subtract $color{#4257b2}xlog 6+2log 5$ from each side to make the terms which contains $color{#4257b2}x$ in the left side alone.

$$
xlog 5+2log 5-xlog 6-2log 5=xlog 6+log 6-xlog 6-2log 5
$$

$$
xlog 5cancel{+2log 5}-xlog 6cancel{-2log 5}=cancel{xlog 6}+log 6cancel{-xlog 6}-2log 5
$$

$$
xlog 5-xlog 6=log 6-2log 5
$$

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

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

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

$$
dfrac{xleft(log 5-log 6right)}{log 5-log 6}=dfrac{log 6-2log 5}{log 5-log 6}
$$

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

Now we can use the calculator to determine $color{#4257b2}log 6$ and $color{#4257b2}log 5$ to find the value of $color{#4257b2}x$.

$$
x=dfrac{0.778-2(0.7)}{0.7-0.778}
$$

$$
x=7.827
$$

So the solution of the equation is $boxed{ x=7.827 }$

Large{$text{color{#c34632}$x=7.827$}$

(a) We would like to solve the equation $color{#4257b2}log_{4} (x+2)+log_{4} (x-1)=1$. First, we note that the left side is a sum of logarithms, so we can use the product law of logarithms where $color{#4257b2}log_{a} x+log_{a} y=log_{a} xy$.

$$
log_{4} (x+2)+log_{4} (x-1)=1
$$

$$
log_{4} (x+2)(x-1)=1
$$

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

Now we have an equation on the logarithmic form, so we can convert it to the exponential form where we know that if $color{#4257b2}log_{a} x=b$, then $color{#4257b2}x=a^{b}$.

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

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

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

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

$$
x^{2}+x-6=0
$$

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

$$
(x-2)(x+3)=0
$$

Now we can use the zero-factor property.

$$
x-2=0 text{or} x+3=0
$$

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

Now we found two solutions for $color{#4257b2}x$, so the next step is to check these values in the original equation to know if they verify it or not.

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

$$
log_{4} (x+2)+log_{4} (x-1)=1
$$

$$
log_{4} (2+2)+log_{4} (2-1)=1
$$

$$
log_{4} 4+log_{4} 1=1
$$

$$
1+0=1
$$

$$
1=1
$$

So the solution $color{#4257b2}x=2$ is true because it verifies the original equation.

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

$$
log_{4} (x+2)+log_{4} (x-1)=1
$$

$$
log_{4} (-3+2)+log_{4} (-3-1)=1
$$

$$
log_{4} (-1)+log_{4} (-4)=1
$$

But we know that $color{#4257b2}log_{4} -1$ and $color{#4257b2}log_{4} -4$ are undefined because the number inside the logarithm must be positive, so the solution $color{#4257b2}x=-3$ is false because it doesn’t verify the original equation.

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

(b) We would like to solve the equation $color{#4257b2}log_{3} (8x-2)+log_{3} (x-1)=2$. First, we note that the left side is a sum of logarithms, so we can use the product law of logarithms where $color{#4257b2}log_{a} x+log_{a} y=log_{a} xy$.

$$
log_{3} (8x-2)+log_{3} (x-1)=2
$$

$$
log_{3} (8x-2)(x-1)=2
$$

$$
log_{3} (8x^{2}-10x+2)=2
$$

Now we have an equation on the logarithmic form, so we can convert it to the exponential form where we know that if $color{#4257b2}log_{a} x=b$, then $color{#4257b2}x=a^{b}$.

$$
8x^{2}-10x+2=3^{2}
$$

$$
8x^{2}-10x+2=9
$$

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

$$
8x^{2}-10x+2-9=0
$$

$$
8x^{2}-10x-7=0
$$

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

$$
8x^{2}-10x-7=0
$$

$$
(4x-7)(2x+1)=0
$$

Now we can use the zero-factor property.

$$
4x-7=0 text{or} 2x+1=0
$$

$$
x=dfrac{7}{4} text{or} x=-dfrac{1}{2}
$$

Now we found two solutions for $color{#4257b2}x$, so the next step is to check these values in the original equation to know if they verify it or not.

For $color{#4257b2}x=dfrac{7}{4}$

$$
log_{3} (8x-2)+log_{3} (x-1)=2
$$

$$
log_{3} left(8(dfrac{7}{4})-2right)+log_{3} left(dfrac{7}{4}-1right)=2
$$

$$
log_{3} 12+log_{3} dfrac{3}{4}=2
$$

$$
2.262+(-0.262)=2
$$

$$
2=2
$$

So the solution $color{#4257b2}x=dfrac{7}{4}$ is true because it verifies the original equation.

For $color{#4257b2}x=-dfrac{1}{2}$

$$
log_{3} (8x-2)+log_{3} (x-1)=2
$$

$$
log_{3} left(8(-dfrac{1}{2})-2right)+log_{3} left(-dfrac{1}{2}-1right)=2
$$

$$
log_{3} (-6)+log_{3} left(-dfrac{3}{2}right)=2
$$

But we know that $color{#4257b2}log_{3} -6$ and $color{#4257b2}log_{3} -dfrac{3}{2}$ are undefined because the number inside the logarithm must be positive, so the solution $color{#4257b2}x=-dfrac{1}{2}$ is false because it doesn’t verify the original equation.

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

Large{$text{color{#c34632}(a) $x=2$ (b) $x=dfrac{7}{4}$}$

Note that the half-life of a radioactive isotope is the time for which the initial amount of isotope decrease by half and that the decay of an isotope is an exponential function. That function we can write as
$$
f(x)=a(1-r)^x.tag{1}
$$
where $a$ is the initial amount of isotope, $r$ is the rate of decay, while $x$ is the number of decay periods.

*(a)* By the definition of half-life, at the end of this period we will have exactly one-half of the initial amount of carbon-14. Since the initial amount is $100 mathrm{g}$, after $5730$ years we will have $50 mathrm{g}$.

*(b)* Note that we need to write Eq$(1)$ for Carbon-14 with a half-life of $5730$ years. The initial amount of carbon is $a=100 mathrm{g}$ while the rate of change from the initial amount to one half is $r=50%=frac{50}{100}=0.5$. The number of half-life periods for time $t$ is
$$
x=dfrac{t}{5730}.$$
Hence, the Eq. $(1)$ now yields
$$
f(t)=100(1-0.5)^tfrac{t}{5730}=100(0.5)^tfrac{t}{5730}.$$

*(c)* The time after which $80 mathrm{g}$ of carbon will remain we can compute by solving the equation
$$
80=100(0.5)^tfrac{t}{5730}.
$$
Dividing by $100$ throughout and reversing the direction of equation, we obtain:
$$
(0.5)^tfrac{t}{5730}=0.8.
$$
Furthermore, we can take logarithm base $0.5$ of both sides and use the property of logarithm $log a^x=xlog a$ and proceed to solve the equation to compute $t$ as follows:
$$
begin{align*}
log_{0.5}(0.5)^tfrac{t}{5730}&=log_{0.5}0.8&&[log a^x=xlog a]\
dfrac{t}{5730}log_{0.5}(0.5)&=log_{0.5}(0.8)&&[log_aa=1]\
dfrac{t}{5730}cdot1&=log_{0.5}(0.8)&&text{[Multiply by $5730$ throughout]}\
t&=5730log_{0.5}(0.8)\
t&approx1844 mathrm{years}.
end{align*}$$

*(d)* The instantaneous rate of change in a decaying carbon we can obtain by choosing a small time interval $[t_1,t_2]$ around the given time $100$ years
and using the formula
$$
r_i=dfrac{f(t_2)-f(t_1)}{t_2-t_1}.$$
Hence, if we choose $t_1=99.9$ and $t_2=100.1$ the amounts of carbon at these points of decay will be
$$
begin{align*}
f(t_1)=100(0.5)^{tfrac{99.9}{5730}}=98.7988\
f(t_2)=100(0.5)^{tfrac{100.1}{5730}}=98.7964.\
end{align*}$$
Using these values for the instantaneous rate of change we obtain
$$
r_i=dfrac{98.7964-98.7988}{100.1-99.9}=-0.012 mathrm{dfrac{g}{year}}.$$

a) $50 mathrm{g}$
b) $f(x)=100(0.5)^tfrac{t}{5730}$
c) $tapprox1844 mathrm{years}$
d) $r_i=-0.012 mathrm{frac{g}{year}}$

*(a)* The cooling time of a hot chocolate from the initial temperature to the given temperature we can obtain by substituting $T=35$ in the given time function. It yields:
$$
begin{align*}
t&=logleft(dfrac{35-22}{75}right):log(0.75)\&=logleft(dfrac{13}{75}right):log(0.75)\&approx6 mathrm{min}.
end{align*}

*(b)* The initial temperature of hot chocolate we can obtain by substituting $t=0$ in the given equation and solving for $T$.
$$
0=logleft(dfrac{T-22}{75}right):log(0.75)implies logleft(dfrac{T-22}{75}right)=0.
$$
Note that $log1=0$, hence we solve for $T$ as follows:
$$begin{align*}
dfrac{T-22}{75}&=1&&text{[Multiply by $75$ throughout]}\
T-22&=75&&text{[Add $22$ throughout]}\
T&=75+22\
T&=97^circmathrm{C}.
end{align*}$$

a) $tapprox6 mathrm{min}$
b) $T=97^circmathrm{C}$