Nelson Functions 11
1st Edition
ISBN: 9780176332037
Textbook solutions
All Solutions
Page 235: Check Your Understanding
Exercise 1
Step 1
1 of 7
a) $x^4(x^3)=x^{4+3}=x^7$
Use the rule:
$$
x^{a} (x^{b})=x^{a+b}
$$
Step 2
2 of 7
b) $(p^{-3})(p)^5=p^{-3+5}=p^2$
Use the rule:
$$
x^{a}( x^{b})=x^{a+b}
$$
Step 3
3 of 7
c) $dfrac{m^5}{m^{-3}}=m^{5-(-3)}=m^8$
Use the rule:
$$
dfrac{x^{a}}{x^{b}}=x^{a-b}
$$
Step 4
4 of 7
d) $dfrac{a^{-4}}{a^{-2}}=a^{-4-(-2)}=a^{-2}=dfrac{1}{a^2}$
Use the rules:
$dfrac{x^{a}}{x^{b}}=x^{a-b}$
$$
x^{-a}=dfrac{1}{x^a}
$$
Step 5
5 of 7
e) $(y^3)^2=y^{3(2)}=y^6$
Use the rule:
$$
(x^{a})^b=x^{ab}
$$
Step 6
6 of 7
f) $(k^6)^{-2}=k^{6(-2)}=k^{-12}=dfrac{1}{k^{12}}$
Use the rule:
$(x^{a})^b=x^{ab}$
$$
x^{-a}=dfrac{1}{x^a}
$$
Result
7 of 7
a) $x^7$
b) $p^2$
c) $m^8$
d) $dfrac{1}{a^2}$
e) $y^6$
f) $dfrac{1}{k^{12}}$
Exercise 2
Step 1
1 of 7
a) $y^{10}(y^4)^{-3}$
$y^{10}(y^4)^{-3}$
$=y^{10}y^{4(-3)}$
$=y^{10}y^{-12}$
$=y^{10+(-12)}$
$=y^{-2}$
$$
=dfrac{1}{y^2}
$$
Use the rules:
$(x^a)^b=x^{ab}$
$$
(x^a)(x^b)=x^{a+b}
$$
Step 2
2 of 7
b) $(x^{-3})^{-3}(x^{-1})^5$
$=x^{-3(-3)}cdot x^{-1(5)}$
$=x^{9}cdot x^{-5}$
$=x^{9-5}$
$$
=x^{4}
$$
Use the rules:
$(x^a)^b=x^{ab}$
$$
(x^a)(x^b)=x^{a+b}
$$
Step 3
3 of 7
c) $dfrac{n^{-4})^3}{(n^{-3})^{-4}}$
$=dfrac{n^{-4(3)}}{n^{-3(-4)}}$
$=dfrac{n^{-12}}{n^{12}}$
$=n^{-12-12}$
$=n^{-24}$
$$
=dfrac{1}{n^{24}}
$$
Use the rules:
$(x^a)^b=x^{ab}$
$dfrac{x^a}{x^b}=x^{a-b}$
$$
x^{-a}=dfrac{1}{x^a}
$$
Step 4
4 of 7
d) $dfrac{w^4(w^{-3})}{(w^{-2})^{-1}}$
$=dfrac{w^{4+(-3)}}{w^{-2(-1)}}$
$=dfrac{w^{1}}{w^2}$
$=w^{1-2}$
$=w^{-1}$
$$
=dfrac{1}{w^{}}
$$
Use the rules:
$(x^a)^b=x^{ab}$
$dfrac{x^a}{x^b}=x^{a-b}$
$$
x^{-a}=dfrac{1}{x^a}
$$
Step 5
5 of 7
e) $dfrac{(x^{-1})^4x}{x^{-3}}$
$=dfrac{x^{-4}cdot x}{x^{-3}}$
$=dfrac{x^{-4+1}}{x^{-3}}$
$=dfrac{x^{-3}}{x^{-3}}$
$=x^{-3-(-3)}$
$=x^0$
$$
=1
$$
Use the rules:
$(x^a)^b=x^{ab}$
$x^acdot x^{b}=x^{a+b}$
$dfrac{x^a}{x^b}=x^{a-b}$
$x^{-a}=dfrac{1}{x^a}$
$x^0=1$ for all $xneq 0$
Step 6
6 of 7
f) $dfrac{(b^{-7})^2}{b(b^{-5})b^9}$
$=dfrac{b^{-7(2)}}{b^{1-5+9}}$
$=dfrac{b^{-14}}{b^5}$
$=b^{-14-5}$
$=b^{-19}$
$$
=dfrac{1}{b^{19}}
$$
Use the rules:
$(x^a)^b=x^{ab}$
$x^acdot x^{b}=x^{a+b}$
$dfrac{x^a}{x^b}=x^{a-b}$
$x^{-a}=dfrac{1}{x^a}$
Result
7 of 7
a) $dfrac{1}{y^2}$
b) $x^4$
c) $dfrac{1}{n^{24}}$
d) $dfrac{1}{w}$
e) $1$
f) $dfrac{1}{b^{19}}$
Exercise 3
Step 1
1 of 2
[begin{gathered}
{text{a) Evaluate the powers}} hfill \
frac{{{x^7}{{left( {{y^2}} right)}^3}}}{{{x^5}{y^4}}} = frac{{{{left( { – 2} right)}^7}{{left( {{3^2}} right)}^3}}}{{{{left( { – 2} right)}^5}{{left( 3 right)}^4}}} hfill \
= frac{{left( { – 128} right)left( {729} right)}}{{left( { – 32} right)left( {81} right)}} hfill \
{text{Multiply}} hfill \
= frac{{ – 93312}}{{ – 2592}} hfill \
{text{Simplify}} hfill \
= 36 hfill \
hfill \
{text{b) Evaluate the powers}} hfill \
frac{{{x^7}{{left( {{y^2}} right)}^3}}}{{{x^5}{y^4}}} = frac{{{x^7}{y^6}}}{{{x^5}{y^4}}} hfill \
= {x^2}{y^2} hfill \
= {left( { – 2} right)^2}{left( 3 right)^2} hfill \
= left( 4 right)left( 9 right) hfill \
{text{Multiply}} hfill \
= 36 hfill \
hfill \
{text{c) A simplified expression is faster to calculate }} hfill \
{text{rather than to substitute the numbers first}}{text{.}} hfill \
hfill \
end{gathered} ]
{text{a) Evaluate the powers}} hfill \
frac{{{x^7}{{left( {{y^2}} right)}^3}}}{{{x^5}{y^4}}} = frac{{{{left( { – 2} right)}^7}{{left( {{3^2}} right)}^3}}}{{{{left( { – 2} right)}^5}{{left( 3 right)}^4}}} hfill \
= frac{{left( { – 128} right)left( {729} right)}}{{left( { – 32} right)left( {81} right)}} hfill \
{text{Multiply}} hfill \
= frac{{ – 93312}}{{ – 2592}} hfill \
{text{Simplify}} hfill \
= 36 hfill \
hfill \
{text{b) Evaluate the powers}} hfill \
frac{{{x^7}{{left( {{y^2}} right)}^3}}}{{{x^5}{y^4}}} = frac{{{x^7}{y^6}}}{{{x^5}{y^4}}} hfill \
= {x^2}{y^2} hfill \
= {left( { – 2} right)^2}{left( 3 right)^2} hfill \
= left( 4 right)left( 9 right) hfill \
{text{Multiply}} hfill \
= 36 hfill \
hfill \
{text{c) A simplified expression is faster to calculate }} hfill \
{text{rather than to substitute the numbers first}}{text{.}} hfill \
hfill \
end{gathered} ]
Result
2 of 2
a) 36
b) $x^2y^2=36$
c) It is often faster to substitute numbers into a simplified form.
Exercise 4
Step 1
1 of 2
[begin{gathered}
{text{a) Evaluate the powers}} hfill \
{left( {p{q^2}} right)^{ – 1}}left( {{p^3}{q^3}} right) = left( {{p^{ – 1}}{q^{ – 2}}} right)left( {{p^3}{q^3}} right) hfill \
= {p^{ – 1 + 3}}{q^{ – 2 + 3}} hfill \
= {p^2}q hfill \
hfill \
{text{b) Evaluate the powers}} hfill \
{left( {frac{{{x^3}}}{y}} right)^{ – 2}} = frac{{{x^{3left( { – 2} right)}}}}{{{y^{ – 2}}}} hfill \
= frac{{{x^{ – 6}}}}{{{y^{ – 2}}}} hfill \
{text{Remember this rule, }}{left( {frac{a}{b}} right)^{ – n}} = {left( {frac{b}{a}} right)^n} hfill \
= frac{{{y^2}}}{{{x^6}}} hfill \
hfill \
{text{c) Evaluate the powers}} hfill \
frac{{{{left( {ab} right)}^{ – 2}}}}{{{b^5}}} = frac{{{a^{ – 2}}{b^{ – 2}}}}{{{b^5}}} hfill \
{text{Simplify the expression}}{text{.}} hfill \
= {a^{ – 2}}{b^{ – 2 – 5}} hfill \
= {a^{ – 2}}{b^{ – 7}} hfill \
{text{Remember this rule, }}{a^{ – n}} = frac{1}{{{a^n}}} hfill \
= frac{1}{{{a^2}{b^7}}} hfill \
hfill \
{text{d) Evaluate the powers}} hfill \
frac{{{m^2}{n^2}}}{{{{left( {{m^3}{n^{ – 2}}} right)}^2}}} = frac{{{m^2}{n^2}}}{{{m^{3left( 2 right)}}{n^{ – 2left( 2 right)}}}} hfill \
= frac{{{m^2}{n^2}}}{{{m^6}{n^4}}} hfill \
{text{Simplfy the expression}}{text{.}} hfill \
= {m^{2 – 6}}{n^{2 – left( { – 4} right)}} hfill \
= {m^{ – 4}}{n^6} hfill \
{text{Remember this rule, }}{left( {frac{a}{b}} right)^{ – n}} = {left( {frac{b}{a}} right)^n} hfill \
= frac{{{m^6}}}{{{n^4}}} hfill \
hfill \
{text{e) }} hfill \
frac{{{{left( {{w^2}x} right)}^2}}}{{{{left( {{x^{ – 1}}} right)}^2}{w^3}}} = frac{{{w^{2left( 2 right)}}{x^2}}}{{{x^{ – 1left( 2 right)}}{w^3}}} hfill \
= frac{{{w^2}{x^2}}}{{{x^{ – 2}}{w^3}}} hfill \
= {w^{2 – 3}}{x^{2 – left( { – 2} right)}} hfill \
= w{x^4} hfill \
hfill \
{text{f)}}{left( {frac{{{{left( {ab} right)}^{ – 1}}}}{{{a^2}{b^{ – 3}}}}} right)^{ – 2}} = {left( {frac{{{a^{ – 1}}{b^{ – 1}}}}{{{a^2}{b^{ – 3}}}}} right)^2} hfill \
= {left( {{a^{ – 1 – 2}}{b^{ – 1 – left( { – 3} right)}}} right)^{ – 2}} hfill \
= {left( {{a^{ – 3}}{b^4}} right)^{ – 2}} hfill \
= left( {{a^6}{b^{ – 4}}} right) hfill \
= frac{{{a^6}}}{{{b^4}}} hfill \
end{gathered} ]
{text{a) Evaluate the powers}} hfill \
{left( {p{q^2}} right)^{ – 1}}left( {{p^3}{q^3}} right) = left( {{p^{ – 1}}{q^{ – 2}}} right)left( {{p^3}{q^3}} right) hfill \
= {p^{ – 1 + 3}}{q^{ – 2 + 3}} hfill \
= {p^2}q hfill \
hfill \
{text{b) Evaluate the powers}} hfill \
{left( {frac{{{x^3}}}{y}} right)^{ – 2}} = frac{{{x^{3left( { – 2} right)}}}}{{{y^{ – 2}}}} hfill \
= frac{{{x^{ – 6}}}}{{{y^{ – 2}}}} hfill \
{text{Remember this rule, }}{left( {frac{a}{b}} right)^{ – n}} = {left( {frac{b}{a}} right)^n} hfill \
= frac{{{y^2}}}{{{x^6}}} hfill \
hfill \
{text{c) Evaluate the powers}} hfill \
frac{{{{left( {ab} right)}^{ – 2}}}}{{{b^5}}} = frac{{{a^{ – 2}}{b^{ – 2}}}}{{{b^5}}} hfill \
{text{Simplify the expression}}{text{.}} hfill \
= {a^{ – 2}}{b^{ – 2 – 5}} hfill \
= {a^{ – 2}}{b^{ – 7}} hfill \
{text{Remember this rule, }}{a^{ – n}} = frac{1}{{{a^n}}} hfill \
= frac{1}{{{a^2}{b^7}}} hfill \
hfill \
{text{d) Evaluate the powers}} hfill \
frac{{{m^2}{n^2}}}{{{{left( {{m^3}{n^{ – 2}}} right)}^2}}} = frac{{{m^2}{n^2}}}{{{m^{3left( 2 right)}}{n^{ – 2left( 2 right)}}}} hfill \
= frac{{{m^2}{n^2}}}{{{m^6}{n^4}}} hfill \
{text{Simplfy the expression}}{text{.}} hfill \
= {m^{2 – 6}}{n^{2 – left( { – 4} right)}} hfill \
= {m^{ – 4}}{n^6} hfill \
{text{Remember this rule, }}{left( {frac{a}{b}} right)^{ – n}} = {left( {frac{b}{a}} right)^n} hfill \
= frac{{{m^6}}}{{{n^4}}} hfill \
hfill \
{text{e) }} hfill \
frac{{{{left( {{w^2}x} right)}^2}}}{{{{left( {{x^{ – 1}}} right)}^2}{w^3}}} = frac{{{w^{2left( 2 right)}}{x^2}}}{{{x^{ – 1left( 2 right)}}{w^3}}} hfill \
= frac{{{w^2}{x^2}}}{{{x^{ – 2}}{w^3}}} hfill \
= {w^{2 – 3}}{x^{2 – left( { – 2} right)}} hfill \
= w{x^4} hfill \
hfill \
{text{f)}}{left( {frac{{{{left( {ab} right)}^{ – 1}}}}{{{a^2}{b^{ – 3}}}}} right)^{ – 2}} = {left( {frac{{{a^{ – 1}}{b^{ – 1}}}}{{{a^2}{b^{ – 3}}}}} right)^2} hfill \
= {left( {{a^{ – 1 – 2}}{b^{ – 1 – left( { – 3} right)}}} right)^{ – 2}} hfill \
= {left( {{a^{ – 3}}{b^4}} right)^{ – 2}} hfill \
= left( {{a^6}{b^{ – 4}}} right) hfill \
= frac{{{a^6}}}{{{b^4}}} hfill \
end{gathered} ]
Result
2 of 2
a) $p^2q$
b) $dfrac{y^2}{x^6}$
c) $dfrac{1}{a^2b^7}$
d) $dfrac{n^6}{m^4}$
e) $wx^4$
f) $dfrac{a^6}{b^4}$
Exercise 5
Step 1
1 of 7
[begin{gathered}
{text{a) }} hfill \
{left( {3x{y^4}} right)^2}{left( {2{x^2}y} right)^3} hfill \
hfill \
= 9{x^2}{y^8} times 8{x^6}{y^3} hfill \
hfill \
= 72{x^8}{y^{11}} hfill \
end{gathered} ]
{text{a) }} hfill \
{left( {3x{y^4}} right)^2}{left( {2{x^2}y} right)^3} hfill \
hfill \
= 9{x^2}{y^8} times 8{x^6}{y^3} hfill \
hfill \
= 72{x^8}{y^{11}} hfill \
end{gathered} ]
Use the rules:
$(x^a)^b=x^{ab}$
$$
x^acdot x^b=x^{a+b}
$$
Step 2
2 of 7
$$
{text{b) }}frac{{{{left( {2{a^3}} right)}^2}}}{{4a{b^2}}} = frac{{4{a^6}}}{{4a{b^2}}} = frac{{{a^5}}}{{{b^2}}}
$$
{text{b) }}frac{{{{left( {2{a^3}} right)}^2}}}{{4a{b^2}}} = frac{{4{a^6}}}{{4a{b^2}}} = frac{{{a^5}}}{{{b^2}}}
$$
Use the rules:
$(x^a)^b=x^{ab}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Step 3
3 of 7
[begin{gathered}
{text{c) }}frac{{{{left( {10x} right)}^{ – 1}} times {y^3}}}{{15{x^3}{y^{ – 3}}}} hfill \
= frac{{{{10}^{ – 1}}{x^{ – 1}}{y^6}}}{{15{x^3}}} hfill \
= frac{{{{10}^{ – 1}}{y^6}}}{{15{x^4}}} hfill \
= frac{{{y^6}}}{{15{x^4} times 10}} hfill \
= frac{{{y^6}}}{{150{x^4}}} hfill \
end{gathered} ]
{text{c) }}frac{{{{left( {10x} right)}^{ – 1}} times {y^3}}}{{15{x^3}{y^{ – 3}}}} hfill \
= frac{{{{10}^{ – 1}}{x^{ – 1}}{y^6}}}{{15{x^3}}} hfill \
= frac{{{{10}^{ – 1}}{y^6}}}{{15{x^4}}} hfill \
= frac{{{y^6}}}{{15{x^4} times 10}} hfill \
= frac{{{y^6}}}{{150{x^4}}} hfill \
end{gathered} ]
Use the rules:
$(x^a)^b=x^{ab}$
$x^{-a}=dfrac{1}{x^a}$
$dfrac{1}{x^{-a}}=x^a$
$dfrac{x^a}{x^b}=x^{a-b}$
Step 4
4 of 7
$$
{text{d) }}frac{{left( {3{m^4}{n^2}} right)}}{{12{m^{ – 2}}{n^6}}} = frac{{9{m^8}{n^4}}}{{12{m^{ – 2}}{n^6}}} = frac{{3{m^{10}}}}{{4{n^2}}}
$$
{text{d) }}frac{{left( {3{m^4}{n^2}} right)}}{{12{m^{ – 2}}{n^6}}} = frac{{9{m^8}{n^4}}}{{12{m^{ – 2}}{n^6}}} = frac{{3{m^{10}}}}{{4{n^2}}}
$$
Use the rules:
$(x^a)^b=x^{ab}$
$x^{-a}=dfrac{1}{x^a}$
$dfrac{1}{x^{-a}}=x^a$
$dfrac{x^a}{x^b}=x^{a-b}$
Step 5
5 of 7
[begin{gathered}
{text{e) }}frac{{{p^{ – 5}} times {{left( {{r^3}} right)}^2}}}{{{{left( {{p^2}r} right)}^2} times {{left( {{p^{ – 1}}} right)}^2}}}{text{ = }}frac{{{p^{ – 5}}{r^6}}}{{{p^4}{r^2}{p^{ – 2}}}} hfill \
hfill \
= frac{{{r^4}}}{{{p^4} times {p^3}}} = frac{{{r^4}}}{{{p^7}}}{text{ }} hfill \
end{gathered} ]
{text{e) }}frac{{{p^{ – 5}} times {{left( {{r^3}} right)}^2}}}{{{{left( {{p^2}r} right)}^2} times {{left( {{p^{ – 1}}} right)}^2}}}{text{ = }}frac{{{p^{ – 5}}{r^6}}}{{{p^4}{r^2}{p^{ – 2}}}} hfill \
hfill \
= frac{{{r^4}}}{{{p^4} times {p^3}}} = frac{{{r^4}}}{{{p^7}}}{text{ }} hfill \
end{gathered} ]
Use the rules:
$(x^a)^b=x^{ab}$
$x^{-a}=dfrac{1}{x^a}$
$dfrac{1}{x^{-a}}=x^a$
$dfrac{x^a}{x^b}=x^{a-b}$
$$
x^atimes x^b=x^{a+b}
$$
Step 6
6 of 7
[begin{gathered}
{text{f) }}{left( {frac{{{{left( {{x^3}y} right)}^{ – 1}} times left( {{x^4}{y^3}} right)}}{{{{left( {{x^2}{y^{ – 3}}} right)}^{ – 2}}}}} right)^{ – 1}} = {left( {frac{{{x^{ – 3}}{y^{ – 1}}{x^4}{y^3}}}{{{x^{ – 4}}{y^6}}}} right)^{ – 1}} hfill \
= {left( {frac{{x{y^{ – 1}}{x^4}}}{{{y^3}}}} right)^{ – 1}} = {left( {frac{{xleft( {{x^4}} right)}}{{{y^4}}}} right)^{ – 1}} hfill \
= {left( {frac{{{x^5}}}{{{y^4}}}} right)^{ – 1}} = frac{{{y^4}}}{{{x^5}}} hfill \
end{gathered} ]
{text{f) }}{left( {frac{{{{left( {{x^3}y} right)}^{ – 1}} times left( {{x^4}{y^3}} right)}}{{{{left( {{x^2}{y^{ – 3}}} right)}^{ – 2}}}}} right)^{ – 1}} = {left( {frac{{{x^{ – 3}}{y^{ – 1}}{x^4}{y^3}}}{{{x^{ – 4}}{y^6}}}} right)^{ – 1}} hfill \
= {left( {frac{{x{y^{ – 1}}{x^4}}}{{{y^3}}}} right)^{ – 1}} = {left( {frac{{xleft( {{x^4}} right)}}{{{y^4}}}} right)^{ – 1}} hfill \
= {left( {frac{{{x^5}}}{{{y^4}}}} right)^{ – 1}} = frac{{{y^4}}}{{{x^5}}} hfill \
end{gathered} ]
Use the rules:
$(x^a)^b=x^{ab}$
$x^{-a}=dfrac{1}{x^a}$
$dfrac{1}{x^{-a}}=x^a$
$dfrac{x^a}{x^b}=x^{a-b}$
$$
x^atimes x^b=x^{a+b}
$$
Result
7 of 7
a) $72x^8y^{11}$
b) $dfrac{a^5}{b^2}$
c) $dfrac{y^6}{150x^4}$
d) $dfrac{3m^{10}}{4n^2}$
e) $dfrac{r^4}{p^7}$
f) $dfrac{y^4}{x^5}$
Exercise 6
Step 1
1 of 7
[begin{gathered}
{text{a) }} hfill \
{left( {{x^4}} right)^{frac{1}{2}}}{left( {{x^6}} right)^{ – frac{1}{3}}} hfill \
= {x^2} cdot {x^{ – 2}} hfill \
= frac{{{x^2}}}{{{x^2}}} hfill \
= 1 hfill \
end{gathered} ]
{text{a) }} hfill \
{left( {{x^4}} right)^{frac{1}{2}}}{left( {{x^6}} right)^{ – frac{1}{3}}} hfill \
= {x^2} cdot {x^{ – 2}} hfill \
= frac{{{x^2}}}{{{x^2}}} hfill \
= 1 hfill \
end{gathered} ]
Use the following rules:
$(x^a)^b=x^{ab}$
$$
x^{-a}=dfrac{1}{x^a}
$$
Step 2
2 of 7
[begin{gathered}
{text{b)}} hfill \
frac{{9{{left( {{c^8}} right)}^{0.5}}}}{{{{left( {16{c^{12}}} right)}^{0.25}}}} = frac{{9{c^2}}}{{{{16}^{0.25}}{c^3}}} hfill \
= frac{{9left( {{c^4}} right)}}{{2{c^3}}} hfill \
= frac{9}{2}{c^{4 – 3}} hfill \
= frac{9}{2}c hfill \
end{gathered} ]
{text{b)}} hfill \
frac{{9{{left( {{c^8}} right)}^{0.5}}}}{{{{left( {16{c^{12}}} right)}^{0.25}}}} = frac{{9{c^2}}}{{{{16}^{0.25}}{c^3}}} hfill \
= frac{{9left( {{c^4}} right)}}{{2{c^3}}} hfill \
= frac{9}{2}{c^{4 – 3}} hfill \
= frac{9}{2}c hfill \
end{gathered} ]
Use the following rules:
$(x^a)^b=x^{ab}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Step 3
3 of 7
[begin{gathered}
{text{c)}} hfill \
frac{{sqrt {25{m^{ – 12}}} }}{{sqrt {36{m^{10}}} }} = frac{{sqrt {25} {m^{ – frac{{12}}{2}}}}}{{sqrt {36} {m^{frac{{10}}{2}}}}} = frac{{5{m^{ – 6}}}}{{6{m^5}}} hfill \
= frac{5}{6}left( {{m^{ – 6 – 5}}} right) = frac{5}{6}{m^{ – 11}} hfill \
= frac{5}{{6{m^{11}}}} hfill \
end{gathered} ]
{text{c)}} hfill \
frac{{sqrt {25{m^{ – 12}}} }}{{sqrt {36{m^{10}}} }} = frac{{sqrt {25} {m^{ – frac{{12}}{2}}}}}{{sqrt {36} {m^{frac{{10}}{2}}}}} = frac{{5{m^{ – 6}}}}{{6{m^5}}} hfill \
= frac{5}{6}left( {{m^{ – 6 – 5}}} right) = frac{5}{6}{m^{ – 11}} hfill \
= frac{5}{{6{m^{11}}}} hfill \
end{gathered} ]
Use the following rules:
$sqrt[m]{a^n}=a^{frac{n}{m}}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Step 4
4 of 7
[begin{gathered}
{text{d)}} hfill \
sqrt[3]{{frac{{{{left( {10{x^3}} right)}^2}}}{{{{left( {10{x^6}} right)}^{ – 1}}}}}} = frac{{{{left( {{{left( {10{x^3}} right)}^2}} right)}^{frac{1}{3}}}}}{{{{left( {{{left( {10{x^6}} right)}^{ – 1}}} right)}^{frac{1}{3}}}}} hfill \
= frac{{{{left( {10{x^3}} right)}^{frac{2}{3}}}}}{{{{left( {10{x^6}} right)}^{frac{{ – 1}}{3}}}}} hfill \
= frac{{{{10}^{frac{2}{3}}}{x^2}}}{{{{10}^{frac{{ – 1}}{3}}}{x^{ – 2}}}} hfill \
= {10^{frac{2}{3} – frac{{ – 1}}{3}}}{x^{2 – left( { – 2} right)}} hfill \
= 10{x^4} hfill \
end{gathered} ]
{text{d)}} hfill \
sqrt[3]{{frac{{{{left( {10{x^3}} right)}^2}}}{{{{left( {10{x^6}} right)}^{ – 1}}}}}} = frac{{{{left( {{{left( {10{x^3}} right)}^2}} right)}^{frac{1}{3}}}}}{{{{left( {{{left( {10{x^6}} right)}^{ – 1}}} right)}^{frac{1}{3}}}}} hfill \
= frac{{{{left( {10{x^3}} right)}^{frac{2}{3}}}}}{{{{left( {10{x^6}} right)}^{frac{{ – 1}}{3}}}}} hfill \
= frac{{{{10}^{frac{2}{3}}}{x^2}}}{{{{10}^{frac{{ – 1}}{3}}}{x^{ – 2}}}} hfill \
= {10^{frac{2}{3} – frac{{ – 1}}{3}}}{x^{2 – left( { – 2} right)}} hfill \
= 10{x^4} hfill \
end{gathered} ]
Use the following rules:
$sqrt[m]{a^n}=a^{frac{n}{m}}$
$(x^a)^b=x^{ab}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Step 5
5 of 7
[begin{gathered}
{text{e)}} hfill \
{left( {frac{{{{left( {32{x^5}} right)}^{ – 2}}}}{{{{left( {{x^{ – 1}}} right)}^{10}}}}} right)^{0.2}} hfill \
= {left( {frac{{{{32}^{ – 2}}{x^{ – 10}}}}{{{x^{ – 10}}}}} right)^{0.2}} hfill \
= {left( {{{32}^{ – 2}}} right)^{0.2}} hfill \
= {32^{ – frac{2}{5}}} hfill \
= frac{1}{{{{32}^{frac{2}{5}}}}} hfill \
= frac{1}{{{{left( {{2^5}} right)}^{frac{2}{5}}}}} hfill \
= frac{1}{{{2^2}}} hfill \
= frac{1}{4} hfill \
end{gathered} ]
{text{e)}} hfill \
{left( {frac{{{{left( {32{x^5}} right)}^{ – 2}}}}{{{{left( {{x^{ – 1}}} right)}^{10}}}}} right)^{0.2}} hfill \
= {left( {frac{{{{32}^{ – 2}}{x^{ – 10}}}}{{{x^{ – 10}}}}} right)^{0.2}} hfill \
= {left( {{{32}^{ – 2}}} right)^{0.2}} hfill \
= {32^{ – frac{2}{5}}} hfill \
= frac{1}{{{{32}^{frac{2}{5}}}}} hfill \
= frac{1}{{{{left( {{2^5}} right)}^{frac{2}{5}}}}} hfill \
= frac{1}{{{2^2}}} hfill \
= frac{1}{4} hfill \
end{gathered} ]
Use the following rules:
$(xy)^a=x^acdot y^a$
$(x^a)^b=x^{ab}$
$x^{-a}=dfrac{1}{x^a}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Step 6
6 of 7
[begin{gathered}
{text{f)}} hfill \
frac{{sqrt[{10}]{{1024{x^{20}}}}}}{{sqrt[9]{{512{x^{27}}}}}} = frac{{sqrt[{10}]{{{2^{10}}{x^{20}}}}}}{{sqrt[9]{{{2^9}{x^{27}}}}}} hfill \
= frac{{2{x^2}}}{{2{x^3}}} = {x^{2 – 3}} hfill \
= {x^{ – 1}} = frac{1}{x} hfill \
end{gathered} ]
{text{f)}} hfill \
frac{{sqrt[{10}]{{1024{x^{20}}}}}}{{sqrt[9]{{512{x^{27}}}}}} = frac{{sqrt[{10}]{{{2^{10}}{x^{20}}}}}}{{sqrt[9]{{{2^9}{x^{27}}}}}} hfill \
= frac{{2{x^2}}}{{2{x^3}}} = {x^{2 – 3}} hfill \
= {x^{ – 1}} = frac{1}{x} hfill \
end{gathered} ]
Use the following rules:
$sqrt[m]{a^n}=a^{frac{n}{m}}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Result
7 of 7
a) 1
b) $dfrac{9c}{2}$
c) $dfrac{5}{6m^{11}}$
d) $10x^4$
e) $dfrac{1}{4}$
f) $dfrac{1}{x}$
Exercise 7
Step 1
1 of 5
[begin{gathered}
{text{a)}} hfill \
{left( {16{x^6}{y^4}} right)^{frac{1}{2}}} = {16^{frac{1}{2}}}{x^{6left( {frac{1}{2}} right)}}{y^{4left( {frac{1}{2}} right)}} hfill \
= 4{x^3}{y^2} hfill \
= 4{left( 2 right)^3}{left( 1 right)^2} hfill \
= 4left( 8 right)left( 1 right) hfill \
= 32 hfill \
end{gathered} ]
{text{a)}} hfill \
{left( {16{x^6}{y^4}} right)^{frac{1}{2}}} = {16^{frac{1}{2}}}{x^{6left( {frac{1}{2}} right)}}{y^{4left( {frac{1}{2}} right)}} hfill \
= 4{x^3}{y^2} hfill \
= 4{left( 2 right)^3}{left( 1 right)^2} hfill \
= 4left( 8 right)left( 1 right) hfill \
= 32 hfill \
end{gathered} ]
Use the rule:
$(x^a)^b=x^{ab}$
Step 2
2 of 5
[begin{gathered}
{text{b)}} hfill \
frac{{{{left( {9{p^{ – 2}}} right)}^{frac{1}{2}}}}}{{6{p^2}}} = frac{{{9^{frac{1}{2}}}{p^{ – 2left( {frac{1}{2}} right)}}}}{{6{p^2}}} hfill \
= frac{{3{p^{ – 1}}}}{{6{p^2}}} hfill \
= frac{3}{6}{p^{ – 1 – 2}} hfill \
= frac{1}{2}{p^{ – 3}} hfill \
= frac{1}{{2{p^3}}} hfill \
= frac{1}{{2{{left( 3 right)}^3}}} hfill \
= frac{1}{{54}} hfill \
end{gathered} ]
{text{b)}} hfill \
frac{{{{left( {9{p^{ – 2}}} right)}^{frac{1}{2}}}}}{{6{p^2}}} = frac{{{9^{frac{1}{2}}}{p^{ – 2left( {frac{1}{2}} right)}}}}{{6{p^2}}} hfill \
= frac{{3{p^{ – 1}}}}{{6{p^2}}} hfill \
= frac{3}{6}{p^{ – 1 – 2}} hfill \
= frac{1}{2}{p^{ – 3}} hfill \
= frac{1}{{2{p^3}}} hfill \
= frac{1}{{2{{left( 3 right)}^3}}} hfill \
= frac{1}{{54}} hfill \
end{gathered} ]
Use the rule:
$(x^a)^b=x^{ab}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Step 3
3 of 5
[begin{gathered}
{text{c) }} hfill \
frac{{{{left( {81{x^4}{y^6}} right)}^{frac{1}{2}}}}}{{8{{left( {{x^9}{y^3}} right)}^{frac{1}{3}}}}} = frac{{{{81}^{^{frac{1}{2}}}}{y^{4left( {frac{1}{2}} right)}}{x^{6left( {frac{1}{2}} right)}}}}{{8left( {{x^{9left( {frac{1}{3}} right)}}{y^{3left( {frac{1}{3}} right)}}} right)}} hfill \
= frac{{9{x^2}{y^3}}}{{8left( {{x^3}y} right)}} hfill \
= frac{9}{8}{x^{2 – 3}}{y^{3 – 1}} hfill \
= frac{9}{8}{x^{ – 1}}{y^2} hfill \
= frac{{9{y^2}}}{{8x}} hfill \
= frac{{9{{left( 5 right)}^2}}}{{8left( {10} right)}} hfill \
= frac{{225}}{{80}}{text{ or }}frac{{45}}{{16}} hfill \
end{gathered} ]
{text{c) }} hfill \
frac{{{{left( {81{x^4}{y^6}} right)}^{frac{1}{2}}}}}{{8{{left( {{x^9}{y^3}} right)}^{frac{1}{3}}}}} = frac{{{{81}^{^{frac{1}{2}}}}{y^{4left( {frac{1}{2}} right)}}{x^{6left( {frac{1}{2}} right)}}}}{{8left( {{x^{9left( {frac{1}{3}} right)}}{y^{3left( {frac{1}{3}} right)}}} right)}} hfill \
= frac{{9{x^2}{y^3}}}{{8left( {{x^3}y} right)}} hfill \
= frac{9}{8}{x^{2 – 3}}{y^{3 – 1}} hfill \
= frac{9}{8}{x^{ – 1}}{y^2} hfill \
= frac{{9{y^2}}}{{8x}} hfill \
= frac{{9{{left( 5 right)}^2}}}{{8left( {10} right)}} hfill \
= frac{{225}}{{80}}{text{ or }}frac{{45}}{{16}} hfill \
end{gathered} ]
Use the rule:
$(x^a)^b=x^{ab}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Step 4
4 of 5
[begin{gathered}
{text{d)}} hfill \
{left( {frac{{{{left( {25{a^4}} right)}^{ – 1}}}}{{{{left( {7{a^{ – 2}}b} right)}^2}}}} right)^{frac{1}{2}}} hfill \
left( {frac{{{{25}^{ – 1}}{a^{4left( { – 1} right)}}}}{{{7^2}{a^{ – 2left( 2 right)}}{b^2}}}} right) hfill \
= {left( {frac{{{{25}^{ – 1}}{a^{ – 4}}}}{{49{a^{ – 4}}{b^2}}}} right)^{frac{1}{2}}} hfill \
= {left( {frac{{{{25}^{ – 1}}}}{{49{b^2}}}} right)^{frac{1}{2}}} hfill \
= {left( {frac{1}{{49{b^2} times 25}}} right)^{frac{1}{2}}} hfill \
= {left( {frac{1}{{1225{b^2}}}} right)^{frac{1}{2}}} hfill \
= {left( {frac{1}{{{{35}^2}{b^2}}}} right)^{frac{1}{2}}} hfill \
= frac{1}{{35b}} = frac{1}{{350}} hfill \
end{gathered} ]
{text{d)}} hfill \
{left( {frac{{{{left( {25{a^4}} right)}^{ – 1}}}}{{{{left( {7{a^{ – 2}}b} right)}^2}}}} right)^{frac{1}{2}}} hfill \
left( {frac{{{{25}^{ – 1}}{a^{4left( { – 1} right)}}}}{{{7^2}{a^{ – 2left( 2 right)}}{b^2}}}} right) hfill \
= {left( {frac{{{{25}^{ – 1}}{a^{ – 4}}}}{{49{a^{ – 4}}{b^2}}}} right)^{frac{1}{2}}} hfill \
= {left( {frac{{{{25}^{ – 1}}}}{{49{b^2}}}} right)^{frac{1}{2}}} hfill \
= {left( {frac{1}{{49{b^2} times 25}}} right)^{frac{1}{2}}} hfill \
= {left( {frac{1}{{1225{b^2}}}} right)^{frac{1}{2}}} hfill \
= {left( {frac{1}{{{{35}^2}{b^2}}}} right)^{frac{1}{2}}} hfill \
= frac{1}{{35b}} = frac{1}{{350}} hfill \
end{gathered} ]
Use the rule:
$(x^a)^b=x^{ab}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Result
5 of 5
a) $4x^3y^2=32$
b) $dfrac{1}{2p^3}=dfrac{1}{54}$
c) $dfrac{9y^2}{8x}=frac{45}{16}$
d) $dfrac{1}{35b}=dfrac{1}{350}$
Exercise 8
Step 1
1 of 5
[begin{gathered}
{text{a)}}{left( {sqrt {10,000x} } right)^{frac{3}{2}}} = {left( {10,000{x^{frac{1}{2}}}} right)^{frac{3}{2}}} hfill \
= {left( {10,000x} right)^{frac{1}{2}left( {frac{3}{2}} right)}} hfill \
= {left( {10,000x} right)^{frac{3}{4}}} hfill \
= sqrt[4]{{10,{{000}^3}{x^{frac{3}{4}}}}} hfill \
= 1000{x^{frac{3}{4}}} hfill \
= 1000{left( {16} right)^{frac{3}{4}}} hfill \
= 1000sqrt[4]{{{{16}^3}}} hfill \
= 1000sqrt[4]{{{{left( {{2^4}} right)}^3}}} hfill \
= 1000left( {{2^3}} right) hfill \
= 8,000 hfill \
end{gathered} ]
{text{a)}}{left( {sqrt {10,000x} } right)^{frac{3}{2}}} = {left( {10,000{x^{frac{1}{2}}}} right)^{frac{3}{2}}} hfill \
= {left( {10,000x} right)^{frac{1}{2}left( {frac{3}{2}} right)}} hfill \
= {left( {10,000x} right)^{frac{3}{4}}} hfill \
= sqrt[4]{{10,{{000}^3}{x^{frac{3}{4}}}}} hfill \
= 1000{x^{frac{3}{4}}} hfill \
= 1000{left( {16} right)^{frac{3}{4}}} hfill \
= 1000sqrt[4]{{{{16}^3}}} hfill \
= 1000sqrt[4]{{{{left( {{2^4}} right)}^3}}} hfill \
= 1000left( {{2^3}} right) hfill \
= 8,000 hfill \
end{gathered} ]
Use the following rules:
$sqrt[m]{a^n}=a^{frac{n}{m}}$
$(x^a)^b=x^{ab}$
Step 2
2 of 5
[begin{gathered}
{text{b)}} hfill \
{left( {frac{{{{left( {4{x^3}} right)}^4}}}{{{{left( {{x^3}} right)}^6}}}} right)^{ – 0.5}} = {left( {frac{{{4^{left( 4 right)}}{x^{3left( 4 right)}}}}{{{x^{3left( 6 right)}}}}} right)^{ – 0.5}} hfill \
= {left( {256{x^{12 – 18}}} right)^{ – 0.5}} hfill \
= {left( {256{x^{ – 6}}} right)^{ – 0.5}} hfill \
= {256^{frac{1}{2}}}{x^3} hfill \
= frac{{{x^3}}}{{16}} hfill \
= frac{{{5^3}}}{{16}} hfill \
= frac{{125}}{{16}} hfill \
end{gathered} ]
{text{b)}} hfill \
{left( {frac{{{{left( {4{x^3}} right)}^4}}}{{{{left( {{x^3}} right)}^6}}}} right)^{ – 0.5}} = {left( {frac{{{4^{left( 4 right)}}{x^{3left( 4 right)}}}}{{{x^{3left( 6 right)}}}}} right)^{ – 0.5}} hfill \
= {left( {256{x^{12 – 18}}} right)^{ – 0.5}} hfill \
= {left( {256{x^{ – 6}}} right)^{ – 0.5}} hfill \
= {256^{frac{1}{2}}}{x^3} hfill \
= frac{{{x^3}}}{{16}} hfill \
= frac{{{5^3}}}{{16}} hfill \
= frac{{125}}{{16}} hfill \
end{gathered} ]
Use the following rules:
$(x^a)^b=x^{ab}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Step 3
3 of 5
[begin{gathered}
{text{c)}}{left( { – 2{a^2}b} right)^{ – 3}}sqrt {25{a^4}{b^6}} hfill \
= left( {{2^{ – 3}}{a^{ – 6}}{b^{ – 3}}} right)left( {{{25}^{frac{1}{2}}}{a^2}{b^3}} right) hfill \
= frac{{ – 5}}{8}{a^{ – 6 + 2}}{b^{ – 3 + 3}} hfill \
= frac{{ – 5}}{8}{a^{ – 4}}{b^0} hfill \
= frac{{ – 5}}{{8{a^4}}} hfill \
= frac{{ – 5}}{{8{{left( 1 right)}^4}}} hfill \
= frac{{ – 5}}{8} hfill \
end{gathered} ]
{text{c)}}{left( { – 2{a^2}b} right)^{ – 3}}sqrt {25{a^4}{b^6}} hfill \
= left( {{2^{ – 3}}{a^{ – 6}}{b^{ – 3}}} right)left( {{{25}^{frac{1}{2}}}{a^2}{b^3}} right) hfill \
= frac{{ – 5}}{8}{a^{ – 6 + 2}}{b^{ – 3 + 3}} hfill \
= frac{{ – 5}}{8}{a^{ – 4}}{b^0} hfill \
= frac{{ – 5}}{{8{a^4}}} hfill \
= frac{{ – 5}}{{8{{left( 1 right)}^4}}} hfill \
= frac{{ – 5}}{8} hfill \
end{gathered} ]
Use the following rules:
$sqrt[m]{a^n}=a^{frac{n}{m}}$
$(x^a)^b=x^{ab}$
$x^atimes x^b=x^{a+b}$
$x^{-a}=dfrac{1}{x^a}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Step 4
4 of 5
[begin{gathered}
{text{d) }} hfill \
sqrt {frac{{left( {18{m^{ – 5}}{n^2}} right)left( {32{m^2}n} right)}}{{4m{n^{ – 3}}}}} = sqrt {frac{{left( {18} right)left( {32} right){m^{ – 5 + 2}}{n^{2 + 1}}}}{{4m{n^{ – 3}}}}} hfill \
= sqrt {left( {frac{{576{m^{ – 3}}n3}}{{4m{n^{ – 3}}}}} right)} hfill \
= {left( {144{m^{ – 3 – 1}}{n^{3 – left( { – 3} right)}}} right)^{frac{1}{2}}} hfill \
= {left( {12{m^{ – 4}}{n^6}} right)^{frac{1}{2}}} hfill \
= left( {{{12}^{^{frac{1}{2}}}}{m^{ – 4left( {frac{1}{2}} right)}}{n^{6left( {frac{1}{2}} right)}}} right) hfill \
= left( {6{m^{ – 2}}{n^3}} right) hfill \
= frac{{6{n^3}}}{{{m^2}}} hfill \
= frac{{6{{left( 1 right)}^3}}}{{{{10}^2}}} hfill \
= frac{{{6^3}}}{{{{10}^2}}} hfill \
= frac{{12}}{{100}}{text{ or }}frac{3}{{25}} hfill \
end{gathered} ]
{text{d) }} hfill \
sqrt {frac{{left( {18{m^{ – 5}}{n^2}} right)left( {32{m^2}n} right)}}{{4m{n^{ – 3}}}}} = sqrt {frac{{left( {18} right)left( {32} right){m^{ – 5 + 2}}{n^{2 + 1}}}}{{4m{n^{ – 3}}}}} hfill \
= sqrt {left( {frac{{576{m^{ – 3}}n3}}{{4m{n^{ – 3}}}}} right)} hfill \
= {left( {144{m^{ – 3 – 1}}{n^{3 – left( { – 3} right)}}} right)^{frac{1}{2}}} hfill \
= {left( {12{m^{ – 4}}{n^6}} right)^{frac{1}{2}}} hfill \
= left( {{{12}^{^{frac{1}{2}}}}{m^{ – 4left( {frac{1}{2}} right)}}{n^{6left( {frac{1}{2}} right)}}} right) hfill \
= left( {6{m^{ – 2}}{n^3}} right) hfill \
= frac{{6{n^3}}}{{{m^2}}} hfill \
= frac{{6{{left( 1 right)}^3}}}{{{{10}^2}}} hfill \
= frac{{{6^3}}}{{{{10}^2}}} hfill \
= frac{{12}}{{100}}{text{ or }}frac{3}{{25}} hfill \
end{gathered} ]
Use the following rules:
$sqrt[m]{a^n}=a^{frac{n}{m}}$
$(x^a)^b=x^{ab}$
$x^atimes x^b=x^{a+b}$
$x^{-a}=dfrac{1}{x^a}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Result
5 of 5
a) $10000x^{frac{3}{4}}=8000$
b) $dfrac{x^3}{16}=dfrac{125}{16}$
c) $-dfrac{5}{8a^4}=-dfrac{5}{8}$
d) $dfrac{12n^3}{m^2}=dfrac{3}{25}$
Exercise 9
Step 1
1 of 5
[begin{gathered}
{text{a)}} hfill \
{left( {36{m^4}{n^6}} right)^{0.5}}{left( {81{m^{12}}{n^8}} right)^{0.25}} hfill \
= {left( {36{m^4}{n^6}} right)^{frac{1}{2}}}{left( {81{m^{12}}{n^8}} right)^{frac{1}{4}}} hfill \
= left( {6{m^2}{n^3}} right)left( {3{m^3}{n^2}} right) hfill \
= 18{m^{2 + 3}}{n^{3 + 2}} hfill \
= 18{m^5}{n^5} hfill \
end{gathered} ]
{text{a)}} hfill \
{left( {36{m^4}{n^6}} right)^{0.5}}{left( {81{m^{12}}{n^8}} right)^{0.25}} hfill \
= {left( {36{m^4}{n^6}} right)^{frac{1}{2}}}{left( {81{m^{12}}{n^8}} right)^{frac{1}{4}}} hfill \
= left( {6{m^2}{n^3}} right)left( {3{m^3}{n^2}} right) hfill \
= 18{m^{2 + 3}}{n^{3 + 2}} hfill \
= 18{m^5}{n^5} hfill \
end{gathered} ]
Use the rules:
$(a^m)^n=a^{mn}$
$$
a^mcdot a^n=a^{m+n}
$$
Step 2
2 of 5
[begin{gathered}
{text{b)}} hfill \
{left( {frac{{{{left( {6{x^3}} right)}^2}left( {6{y^3}} right)}}{{{{left( {9xy} right)}^6}}}} right)^{frac{{ – 1}}{3}}} = {left( {frac{{36 times 6}}{{{9^6}{y^3}}}} right)^{ – frac{1}{3}}} hfill \
= {left( {frac{{{9^6}{y^3}}}{{36 times 6}}} right)^{frac{1}{3}}} hfill \
= {left( {frac{{{9^6}{y^3}}}{{216}}} right)^{frac{1}{3}}} hfill \
= {left( {frac{{{9^6}{y^3}}}{{{6^3}}}} right)^{frac{1}{3}}} hfill \
= frac{{{9^2}y}}{6} hfill \
= frac{{81y}}{6}{text{ or }}frac{{27y}}{2} hfill \
end{gathered} ]
{text{b)}} hfill \
{left( {frac{{{{left( {6{x^3}} right)}^2}left( {6{y^3}} right)}}{{{{left( {9xy} right)}^6}}}} right)^{frac{{ – 1}}{3}}} = {left( {frac{{36 times 6}}{{{9^6}{y^3}}}} right)^{ – frac{1}{3}}} hfill \
= {left( {frac{{{9^6}{y^3}}}{{36 times 6}}} right)^{frac{1}{3}}} hfill \
= {left( {frac{{{9^6}{y^3}}}{{216}}} right)^{frac{1}{3}}} hfill \
= {left( {frac{{{9^6}{y^3}}}{{{6^3}}}} right)^{frac{1}{3}}} hfill \
= frac{{{9^2}y}}{6} hfill \
= frac{{81y}}{6}{text{ or }}frac{{27y}}{2} hfill \
end{gathered} ]
Use the rules:
$(a^m)^n=a^{mn}$
$a^mcdot a^n=a^{m+n}$
$$
dfrac{a^m}{a^n}=a^{m-n}
$$
Step 3
3 of 5
[begin{gathered}
{text{c)}} hfill \
{left( {frac{{sqrt {64{a^{12}}} }}{{{{left( {{a^{1.5}}} right)}^{ – 6}}}}} right)^{frac{2}{3}}} = {left( {frac{{8{a^6}}}{{{a^{ – 9}}}}} right)^{frac{2}{3}}} hfill \
= {left( {8{a^{15}}} right)^{frac{2}{3}}} hfill \
= sqrt[3]{{{{left( {8{a^{15}}} right)}^2}}} hfill \
= sqrt[3]{{64{a^{30}}}} hfill \
= 4{a^{10}} hfill \
end{gathered} ]
{text{c)}} hfill \
{left( {frac{{sqrt {64{a^{12}}} }}{{{{left( {{a^{1.5}}} right)}^{ – 6}}}}} right)^{frac{2}{3}}} = {left( {frac{{8{a^6}}}{{{a^{ – 9}}}}} right)^{frac{2}{3}}} hfill \
= {left( {8{a^{15}}} right)^{frac{2}{3}}} hfill \
= sqrt[3]{{{{left( {8{a^{15}}} right)}^2}}} hfill \
= sqrt[3]{{64{a^{30}}}} hfill \
= 4{a^{10}} hfill \
end{gathered} ]
Use the rules:
$(a^m)^n=a^{mn}$
$sqrt[m]{a^n}=a^{frac{n}{m}}$
$a^mcdot a^n=a^{m+n}$
$$
dfrac{a^m}{a^n}=a^{m-n}
$$
Step 4
4 of 5
[begin{gathered}
{text{d)}}{left( {frac{{{{left( {{x^{18}}} right)}^{frac{{ – 1}}{6}}}}}{{sqrt[5]{{243}}{x^{10}}}}} right)^{0.5}} = {left( {frac{{{x^{ – 3}}}}{{{{243}^{frac{1}{5}}}{x^2}}}} right)^{0.5}} hfill \
= {left( {frac{{{x^{ – 3 – 2}}}}{3}} right)^{0.5}} hfill \
= frac{{{x^{ – 5left( {0.5} right)}}}}{{{3^{0.5}}}} hfill \
= frac{{{x^{ – frac{5}{2}}}}}{{sqrt 3 }} hfill \
= frac{1}{{sqrt 3 {x^{frac{5}{2}}}}} hfill \
end{gathered} ]
{text{d)}}{left( {frac{{{{left( {{x^{18}}} right)}^{frac{{ – 1}}{6}}}}}{{sqrt[5]{{243}}{x^{10}}}}} right)^{0.5}} = {left( {frac{{{x^{ – 3}}}}{{{{243}^{frac{1}{5}}}{x^2}}}} right)^{0.5}} hfill \
= {left( {frac{{{x^{ – 3 – 2}}}}{3}} right)^{0.5}} hfill \
= frac{{{x^{ – 5left( {0.5} right)}}}}{{{3^{0.5}}}} hfill \
= frac{{{x^{ – frac{5}{2}}}}}{{sqrt 3 }} hfill \
= frac{1}{{sqrt 3 {x^{frac{5}{2}}}}} hfill \
end{gathered} ]
Use the rules:
$(a^m)^n=a^{mn}$
$sqrt[m]{a^n}=a^{frac{n}{m}}$
$a^mcdot a^n=a^{m+n}$
$dfrac{a^m}{a^n}=a^{m-n}$
$$
a^{-n}=dfrac{1}{a^n}
$$
Result
5 of 5
a) $18m^5n^5$
b) $dfrac{27y}{2}$
c) $4a^{10}$
d) $dfrac{1}{sqrt{3}x^{frac{5}{2}}}$
Exercise 10
Step 1
1 of 3
[begin{gathered}
{text{Simplify the expression by applying rules of exponents}} hfill \
hfill \
frac{{{{left( {16{x^8}{y^{ – 4}}} right)}^{frac{1}{4}}}}}{{32{x^{ – 2}}{y^8}}} = frac{{{{16}^{frac{1}{4}}}{x^2}{y^{ – 1}}}}{{32{x^{ – 2}}{y^8}}} hfill \
hfill \
= frac{2}{{32}}left( {{x^{2 – left( { – 2} right)}}{y^{ – 1 – 8}}} right) hfill \
hfill \
= frac{1}{{16}}left( {{x^4}{y^9}} right) hfill \
hfill \
= frac{{{x^4}}}{{16{y^9}}} hfill \
end{gathered} ]
{text{Simplify the expression by applying rules of exponents}} hfill \
hfill \
frac{{{{left( {16{x^8}{y^{ – 4}}} right)}^{frac{1}{4}}}}}{{32{x^{ – 2}}{y^8}}} = frac{{{{16}^{frac{1}{4}}}{x^2}{y^{ – 1}}}}{{32{x^{ – 2}}{y^8}}} hfill \
hfill \
= frac{2}{{32}}left( {{x^{2 – left( { – 2} right)}}{y^{ – 1 – 8}}} right) hfill \
hfill \
= frac{1}{{16}}left( {{x^4}{y^9}} right) hfill \
hfill \
= frac{{{x^4}}}{{16{y^9}}} hfill \
end{gathered} ]
Step 2
2 of 3
[begin{gathered}
{text{a) If }}M = 1 hfill \
frac{{{x^4}}}{{16{y^9}}} = 1 hfill \
{x^4} = 16{y^9} hfill \
{text{By inspection, we see that a solution is}} hfill \
x = 2{text{ and }}y = 1{text{ because}} hfill \
{2^4} = 16 hfill \
hfill \
{text{b)}};{text{If }}M > 1 hfill \
{text{ }}frac{{{x^4}}}{{16{y^9}}} > 1 hfill \
{text{Based on part}}left( {text{a}} right),{text{ we can expect that}} hfill \
{text{this happens when }}x > 2{text{ and/or }},{text{0}} < y leqslant 1 hfill \
{text{An example would be}} hfill \
x = 3{text{ and }}y = 1 hfill \
hfill \
{text{c) 0}} < M < 1 hfill \
0 < frac{{{x^4}}}{{16{y^9}}} < 1 hfill \
{text{This happens when }}0 < x < 2{text{ while }}y = 1 hfill \
{text{An example would be}} hfill \
x = 1{text{ and }}y = 1 hfill \
hfill \
{text{d) }}M < 0 hfill \
frac{{{x^4}}}{{16{y^9}}} < 0 hfill \
{x^4}{text{ has even exponent so it is always positive}}{text{. To}} hfill \
{text{make }}M{text{ negative, }}y{text{ must be negative}} hfill \
hfill \
{text{An example would be}} hfill \
x = 3{text{ and }}y = – 1 hfill \
end{gathered} ]
{text{a) If }}M = 1 hfill \
frac{{{x^4}}}{{16{y^9}}} = 1 hfill \
{x^4} = 16{y^9} hfill \
{text{By inspection, we see that a solution is}} hfill \
x = 2{text{ and }}y = 1{text{ because}} hfill \
{2^4} = 16 hfill \
hfill \
{text{b)}};{text{If }}M > 1 hfill \
{text{ }}frac{{{x^4}}}{{16{y^9}}} > 1 hfill \
{text{Based on part}}left( {text{a}} right),{text{ we can expect that}} hfill \
{text{this happens when }}x > 2{text{ and/or }},{text{0}} < y leqslant 1 hfill \
{text{An example would be}} hfill \
x = 3{text{ and }}y = 1 hfill \
hfill \
{text{c) 0}} < M < 1 hfill \
0 < frac{{{x^4}}}{{16{y^9}}} < 1 hfill \
{text{This happens when }}0 < x < 2{text{ while }}y = 1 hfill \
{text{An example would be}} hfill \
x = 1{text{ and }}y = 1 hfill \
hfill \
{text{d) }}M < 0 hfill \
frac{{{x^4}}}{{16{y^9}}} < 0 hfill \
{x^4}{text{ has even exponent so it is always positive}}{text{. To}} hfill \
{text{make }}M{text{ negative, }}y{text{ must be negative}} hfill \
hfill \
{text{An example would be}} hfill \
x = 3{text{ and }}y = – 1 hfill \
end{gathered} ]
Result
3 of 3
$M=dfrac{x^4}{16y^9}$
a) $x=2$ , $y=1$
b) $x=3$, $y=1$
c) $x=1$, $y=1$
d) $x=3$ , $y=-1$
Exercise 11
Step 1
1 of 2
[begin{gathered}
{text{a) The surface area to volume ratio is }}frac{{SA}}{V} hfill \
frac{{SA}}{V} = frac{{2pi rh + 2pi {r^2}}}{{pi {r^2}h}} hfill \
hfill \
= frac{{pi rleft( {2h + 2r} right)}}{{pi rleft( {rh} right)}} hfill \
hfill \
= frac{{left( {2h + 2r} right)}}{{rh}} hfill \
hfill \
{text{b) when }}r = 0.8{text{ cm and }}h = 12{text{ cm}} hfill \
hfill \
frac{{left( {2h + 2r} right)}}{{rh}} = frac{{left( {2left( {12} right) + 2left( {0.8} right)} right)}}{{left( {0.8} right)left( {12} right)}} hfill \
hfill \
= frac{{left( {24 + 1.6} right)}}{{9.6}} hfill \
hfill \
= frac{{8{text{ cm}}}}{{3{text{ c}}{{text{m}}^{text{2}}}}}{text{ or }}2.67{text{ c}}{{text{m}}^{ – 1}} hfill \
end{gathered} ]
{text{a) The surface area to volume ratio is }}frac{{SA}}{V} hfill \
frac{{SA}}{V} = frac{{2pi rh + 2pi {r^2}}}{{pi {r^2}h}} hfill \
hfill \
= frac{{pi rleft( {2h + 2r} right)}}{{pi rleft( {rh} right)}} hfill \
hfill \
= frac{{left( {2h + 2r} right)}}{{rh}} hfill \
hfill \
{text{b) when }}r = 0.8{text{ cm and }}h = 12{text{ cm}} hfill \
hfill \
frac{{left( {2h + 2r} right)}}{{rh}} = frac{{left( {2left( {12} right) + 2left( {0.8} right)} right)}}{{left( {0.8} right)left( {12} right)}} hfill \
hfill \
= frac{{left( {24 + 1.6} right)}}{{9.6}} hfill \
hfill \
= frac{{8{text{ cm}}}}{{3{text{ c}}{{text{m}}^{text{2}}}}}{text{ or }}2.67{text{ c}}{{text{m}}^{ – 1}} hfill \
end{gathered} ]
Result
2 of 2
a) $dfrac{2h+2r}{hr}$
b) $dfrac{SA}{V}=2.67;text{cm}^{-1}$
Exercise 12
Step 1
1 of 2
[begin{gathered}
{text{We shall simplify the expressions first}} hfill \
frac{{{{text{y}}^{ – 4}}{{left( {{x^2}} right)}^{ – 3}}{y^{ – 3}}}}{{{x^{ – 5}}{{left( {{y^{ – 4}}} right)}^2}}} = frac{{{y^{ – 4}}left( {{x^{ – 6}}} right){y^{ – 3}}}}{{{x^{ – 5}}{y^{ – 8}}}} hfill \
= frac{{{y^{ – 7}}{x^{ – 6}}}}{{{x^{ – 5}}{y^{ – 8}}}} hfill \
= {y^{ – 7 – 8}}{x^{ – 6 – 5}} hfill \
= frac{y}{x} hfill \
{text{Substitute the given values}} hfill \
frac{y}{x} = frac{{ – 3}}{2} hfill \
hfill \
{text{We shall simplify the expressions first}} hfill \
frac{{{x^{ – 3}}{{left( {{y^{ – 1}}} right)}^2}}}{{left( {{x^{ – 5}}} right)left( {{y^4}} right)}} = frac{{{x^{ – 3}}{y^{ – 2}}}}{{{x^{ – 5}}{y^4}}} hfill \
= {x^{ – 3 – left( { – 5} right)}}{y^{ – 2 – 4}} hfill \
= {x^2}{y^{ – 6}} hfill \
= frac{{{x^2}}}{{{y^6}}} hfill \
{text{Substitute the given values}} hfill \
frac{{{x^2}}}{{{y^6}}} = frac{{{2^2}}}{{{3^6}}} hfill \
= frac{4}{729} hfill \
hfill \
{text{We shall simplify the expressions first}} hfill \
left( {{y^{ – 5}}} right){left( {{x^5}} right)^{ – 2}}left( {{y^2}} right){left( {{x^{ – 3}}} right)^{ – 4}} = {y^{ – 5 + 2}}{x^{ – 10 + 12}} hfill \
= {y^{ – 3}}{x^2} hfill \
= frac{{{x^2}}}{{{y^3}}} hfill \
{text{Substitute the given values}} hfill \
= frac{{{2^2}}}{{{3^3}}} hfill \
= frac{4}{{27}} hfill \
hfill \
{text{The order from least to greatest is}} hfill \
frac{{ – 3}}{2},frac{4}{{27}},frac{4}{9} hfill \
end{gathered} ]
{text{We shall simplify the expressions first}} hfill \
frac{{{{text{y}}^{ – 4}}{{left( {{x^2}} right)}^{ – 3}}{y^{ – 3}}}}{{{x^{ – 5}}{{left( {{y^{ – 4}}} right)}^2}}} = frac{{{y^{ – 4}}left( {{x^{ – 6}}} right){y^{ – 3}}}}{{{x^{ – 5}}{y^{ – 8}}}} hfill \
= frac{{{y^{ – 7}}{x^{ – 6}}}}{{{x^{ – 5}}{y^{ – 8}}}} hfill \
= {y^{ – 7 – 8}}{x^{ – 6 – 5}} hfill \
= frac{y}{x} hfill \
{text{Substitute the given values}} hfill \
frac{y}{x} = frac{{ – 3}}{2} hfill \
hfill \
{text{We shall simplify the expressions first}} hfill \
frac{{{x^{ – 3}}{{left( {{y^{ – 1}}} right)}^2}}}{{left( {{x^{ – 5}}} right)left( {{y^4}} right)}} = frac{{{x^{ – 3}}{y^{ – 2}}}}{{{x^{ – 5}}{y^4}}} hfill \
= {x^{ – 3 – left( { – 5} right)}}{y^{ – 2 – 4}} hfill \
= {x^2}{y^{ – 6}} hfill \
= frac{{{x^2}}}{{{y^6}}} hfill \
{text{Substitute the given values}} hfill \
frac{{{x^2}}}{{{y^6}}} = frac{{{2^2}}}{{{3^6}}} hfill \
= frac{4}{729} hfill \
hfill \
{text{We shall simplify the expressions first}} hfill \
left( {{y^{ – 5}}} right){left( {{x^5}} right)^{ – 2}}left( {{y^2}} right){left( {{x^{ – 3}}} right)^{ – 4}} = {y^{ – 5 + 2}}{x^{ – 10 + 12}} hfill \
= {y^{ – 3}}{x^2} hfill \
= frac{{{x^2}}}{{{y^3}}} hfill \
{text{Substitute the given values}} hfill \
= frac{{{2^2}}}{{{3^3}}} hfill \
= frac{4}{{27}} hfill \
hfill \
{text{The order from least to greatest is}} hfill \
frac{{ – 3}}{2},frac{4}{{27}},frac{4}{9} hfill \
end{gathered} ]
Result
2 of 2
[begin{gathered}
{text{The order from least to greatest is}} hfill \
frac{{ – 3}}{2},frac{4}{{27}},frac{4}{729} hfill \
end{gathered} ]
{text{The order from least to greatest is}} hfill \
frac{{ – 3}}{2},frac{4}{{27}},frac{4}{729} hfill \
end{gathered} ]
Exercise 13
Step 1
1 of 3
$$
bold{Similarities}
$$
bold{Similarities}
$$
$$
bold{Differences}
$$
bold{Differences}
$$
Step 2
2 of 3
Both processes need to follow the order of operations.
(1) terms inside parenthesis
(2) exponent
(3) multiplication or division (whichever comes first)
(4) addition or subtraction (whichever comes first)
In simplifying algebraic expression, you can add or subtract similar terms (you cannot add unlike terms), however, you can multiply or divide terms regardless of whether they are similar or not.
Result
3 of 3
Both processes need to follow the order of operations. The difference is that for algebraic expression, you can only add or subtract similar terms.
Exercise 14
Step 1
1 of 2
[begin{gathered}
{text{a) We need to express the radius }}r{text{ in terms of the volume }}V hfill \
V = frac{4}{3}pi {r^3} hfill \
{r^3} = frac{V}{{frac{4}{3}pi }} hfill \
{r^3} = frac{{3V}}{{4pi }} hfill \
r = sqrt[3]{{frac{{3V}}{{4pi }}}}{text{ or }}r = {left( {frac{{3V}}{{4pi }}} right)^{frac{1}{3}}} hfill \
hfill \
{text{b) We need to find }}r{text{ when }}V = frac{{256}}{3}{{text{m}}^{text{3}}} hfill \
r = {left( {frac{{3V}}{{4pi }}} right)^{frac{1}{3}}} hfill \
= {left( {frac{{3left( {frac{{256}}{3}{{text{m}}^{text{3}}}} right)}}{{4pi }}} right)^{frac{1}{3}}} hfill \
= {left( {frac{{256}}{4}{{text{m}}^{text{3}}}} right)^{frac{1}{3}}} hfill \
= {left( {64{{text{m}}^{text{3}}}} right)^{frac{1}{3}}} hfill \
= sqrt[3]{{64{{text{m}}^{text{3}}}}} hfill \
r = 4{text{ m}} hfill \
hfill \
hfill \
end{gathered} ]
{text{a) We need to express the radius }}r{text{ in terms of the volume }}V hfill \
V = frac{4}{3}pi {r^3} hfill \
{r^3} = frac{V}{{frac{4}{3}pi }} hfill \
{r^3} = frac{{3V}}{{4pi }} hfill \
r = sqrt[3]{{frac{{3V}}{{4pi }}}}{text{ or }}r = {left( {frac{{3V}}{{4pi }}} right)^{frac{1}{3}}} hfill \
hfill \
{text{b) We need to find }}r{text{ when }}V = frac{{256}}{3}{{text{m}}^{text{3}}} hfill \
r = {left( {frac{{3V}}{{4pi }}} right)^{frac{1}{3}}} hfill \
= {left( {frac{{3left( {frac{{256}}{3}{{text{m}}^{text{3}}}} right)}}{{4pi }}} right)^{frac{1}{3}}} hfill \
= {left( {frac{{256}}{4}{{text{m}}^{text{3}}}} right)^{frac{1}{3}}} hfill \
= {left( {64{{text{m}}^{text{3}}}} right)^{frac{1}{3}}} hfill \
= sqrt[3]{{64{{text{m}}^{text{3}}}}} hfill \
r = 4{text{ m}} hfill \
hfill \
hfill \
end{gathered} ]
Result
2 of 2
a) $r=sqrt[3]{dfrac{3V}{4pi}}=left(dfrac{3V}{4pi}right)^{frac{1}{3}}$
b) $r=4$ m
Exercise 15
Step 1
1 of 3
Remember that
$x^acdot x^b=x^{a+b}$
$sqrt[m]{a^m}=a$
$(x^a)^b=x^{ab}$
$$
dfrac{x^a}{x^b}=x^{a-b}
$$
Step 2
2 of 3
$$
frac{{sqrt {xleft( {{x^{2n + 1}}} right)} }}{{sqrt[3]{{{x^{3n}}}}}} = frac{{sqrt {{x^{2n + 2}}} }}{{{x^n}}} = frac{{{{left( {{x^{2n + 2}}} right)}^{frac{1}{2}}}}}{{{x^n}}} = frac{{{x^{n + 1}}}}{{{x^n}}} = {x^{n + 1 – n}} = x
$$
frac{{sqrt {xleft( {{x^{2n + 1}}} right)} }}{{sqrt[3]{{{x^{3n}}}}}} = frac{{sqrt {{x^{2n + 2}}} }}{{{x^n}}} = frac{{{{left( {{x^{2n + 2}}} right)}^{frac{1}{2}}}}}{{{x^n}}} = frac{{{x^{n + 1}}}}{{{x^n}}} = {x^{n + 1 – n}} = x
$$
Result
3 of 3
$$
x
$$
x
$$
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