Question 1: Simplify:

$a^6\times{}a^8=\ a^{14}$

$x^5\times{}x^{-3}=x^2$

$z^9\times{}z^3\times{}z^{-6}=z^6$

$a^2b^3\times{}a^5b^2=a^7b^5$

$5x^7\times{}3x^4=15x^{11}$

$p^3q^4\times{}p^5q^{-5}=p^8q^{-1}$

$x^7y^{-5}\times{}x^{-5}y^3=x^2y^{-2}$

$x^{-2}y^5\times{}x^0y^{-7}=x^{-2}y^{-2}$

$x^6y^4z^{-2}\times{}x^{-3}y^{-5}z^{-1}\times{}x^2y^0z^4=x^5y^{-1}z^1$

$\frac{x^{12}}{x^7}$ $=\ x^5$

$\frac{z^6}{z^{-3}}$ $=\ =\ z^9$

$\frac{m^5}{m^2}\frac{n^3}{n^{-4}}$ $=\ m^3n^7$

$\frac{18x^9}{6x^7}$ $=3x^2$

$\frac{7a^{12}}{56a^{15}}=\frac{a^{-3}}{8}$

$\frac{a^{13}}{a^5}\frac{b^7}{b^{-3}}$ $=a^8b^{10}$

$\frac{7x^{14}}{21x^{-10}}=\frac{x^{24}}{3}$

$\frac{p^{11}}{p^{11}}$ $=\ p^0=1$

$\frac{a^7}{a^{-2}}\frac{b^5}{b^3}\frac{c^4}{c^6}$ $=\ a^9b^2c^{-2}$

${(a^3)}^2=\ a^6$

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

${(2x^2y)}^4=16x^8y^4$

${(3x^3y^{-3})}^2=9x^6y^{-6}$

${(m^2n^{-3})}^4=m^8n^{-12}$

Question 2: Evaluate

${(36)}^{1/2}={(6^2)}^{1/2}=6$

${(64)}^{-1/3}={(4^3)}^{-1/3}=\frac{1}{4}$

${(27)}^{2/3}={(3^3)}^{2/3}=9$

${(81)}^{-1/4}={(3^4)}^{-1/4}=$ $\frac{1}{3}$

${(16)}^{-3/4}={(2^4)}^{-3/4}=$ $\frac{1}{8}$

${(32)}^{-4/5}={(2^5)}^{-4/5}=$ $\frac{1}{16}$

Question 3: Simplify

${(25a^2)}^{1/2}={(5^2a^2)}^{1/2}=5a$

${(27x^{-3})}^{1/3}={(3^3x^{-3})}^{1/3}=\frac{3}{x}$

${(64m^{-6}n^3)}^{2/3}=(4^3{m^{-6}n^3)}^{\frac{2}{3}}=16m^{-4}n^2$

${(81a^4b^8c^{-4})}^{1/4}=3ab^2c^{-1}$

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

${(6ab^2c^{-3})}^{-1}=\ \frac{c^3}{6ab^2}$

${(-3a^{3/4}b^{-1/4})}^4=\ \frac{{-a}^3}{3b}$

${(32a^{10}b^{-5})}^{1/5}=\ {(2^5a^{10}b^{-5})}^{1/5}=2a^2b^{-1}$

$\sqrt[3]{x^{18}y^{-12}z^3}=({x^{18}y^{-12}z^3)}^{\frac{1}{3}}=\ x^6y^{-4}z$

Question 4: Show that:

(i) $\frac{x^{m+n}x^{n+l}x^{l+m}}{{(x^mx^nx^l)}^2}$ $=1$

$\frac{x^{2m+2n+2l}}{{(x^mx^nx^l)}^2}$ $=1$

$\frac{x^{2m}x^{2n}x^{2l}}{{(x^mx^nx^l)}^2}$ $=1$

$\frac{{{(x}^mx^nx^l)}^2}{{(x^mx^nx^l)}^2}$ $=1$

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(ii) $\sqrt{x^{p-q}}\sqrt{x^{q-r}}\sqrt{x^{r-p}}$ $=1$

${(x^{p-q+q-r+r-p})}^{1/2}=\ {x^0}^{1/2}=\ x^0=1$

${\left(\frac{x^p}{x^q}\right)}^r\times{}{\left(\frac{x^q}{x^r}\right)}^p\times{} {\left(\frac{x^r}{x^p}\right)}^q=1$

${{(x}^{p-q})}^r{{(x}^{q-r})}^p{{(x}^{r-p})}^q=1$

$x^{pr-qr+pq-rp+rq-pq}=1\$

$x^0=1\ (Hence\ Proved)$

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(iii) ${{(x}^{a+b})}^{\left(a-b\right)}({x^{b+c})}^{\left(b-c\right)}{{(x}^{c+a})}^{\left(c-a\right)}=\ 1$

$x^{a^2-b^2}x^{b^2-c^2}x^{c^2-a^2}=1$

$x^0=1\ Hence\ Proved$

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(iv) ${\left(\frac{x^a}{x^{-b}}\right)}^{a-b}\times{}{\left(\frac{x^b}{x^{-c}}\right)}^{b-c}\times{}{\left(\frac{x^c}{x^{-a}}\right)}^{c-a}$ $=1$

${\left(x^{a+b}\right)}^{a-b}\times{}{\left(x^{b+c}\right)}^{b-c}\times{}{\left(x^{c+a}\right)}^{c-a}=1$

$x^{a^2-b^2}x^{b^2-c^2}x^{c^2-a^2}=1$

$x^0=1\ Hence\ Proved$

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(v) ${\left(\frac{x^{a+b}}{x^c}\right)}^{a-b}\times{}{\left(\frac{x^{b+c}}{x^a}\right)}^{b-c}\times{}{\left(\frac{x^{c+a}}{x^b}\right)}^{c-a}$ $=1$

${\left(x^{a+b-c}\right)}^{a-b}\times{}{\left(x^{b+c-a}\right)}^{b-c}\times{}{\left(x^{c+a-b}\right)}^{c-a}=1$

$x^{a^2+ab-ac-ab-b^2+bc+b^2+bc-ab-bc-c^2+ac+c^2+ac-bc-ac-a^2+ab}=1$

$x^0=1\ Hence\ Proved$

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(vi) ${\left(\frac{x^{a^2}}{x^{b^2}}\right)}^{\frac{1}{a+b}}{\times{}\left(\frac{x^{b^2}}{x^{c^2}}\right)}^{\frac{1}{b+c}}\times{}{\left(\frac{x^{c^2}}{x^{a^2}}\right)}^{\frac{1}{c+a}}$ $=1$

${\left(x^{a^2-b^2}\right)}^{\frac{1}{a+b}}\times{}{\left(x^{b^2-c^2}\right)}^{\frac{1}{b+c}}\times{}{\left(x^{c^2-a^2}\right)}^{\frac{1}{c+a}}=1$

$x^{\frac{(a+b)(a-b)}{(a+b)}}{\times{}x}^{\frac{(b+c)(b-c)}{(b+c)}}\times{}x^{\frac{(c+a)(c-a)}{(c+a)}}=1$

$x^{a-b}{\times{}x}^{b-c}\times{}x^{c-a}=1$

$x^0=1\ Hence\ Proved$

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(vii) ${\left(\frac{x^a}{x^b}\right)}^{a^2+ab+b^2}\times{}{\left(\frac{x^b}{x^c}\right)}^{b^2+bc+c^2}\times{}{\left(\frac{x^c}{x^a}\right)}^{c^2+ca+a^2}$ $=1$

${\left(x^{a-b}\right)}^{a^2+ab+b^2}\times{}{\left(x^{b-c}\right)}^{b^2+bc+c^2}\times{}{\left(x^{c-a}\right)}^{c^2+ca+a^2}=1$

$\left(x^{a^3-b^3}\right)\left(x^{b^3-c^3}\right)\left(x^{c^3-a^3}\right)=1$

$x^0=1\ Hence\ Proved$

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Question 5: Evaluate

i) ${\left(\frac{x^a}{x^b}\right)}^{\frac{1}{ab}}{\times{}\left(\frac{x^b}{x^c}\right)}^{\frac{1}{bc}}\times{}{\left(\frac{x^c}{x^a}\right)}^{\frac{1}{ca}}$

$= x^{\frac{a-b}{ab}}x^{\frac{b-c}{bc}}x^{\frac{c-a}{ca}}$

$= x^{\frac{1}{b}-\frac{1}{a}+\frac{1}{c}-\frac{1}{b}+\frac{1}{a}-\frac{1}{c}}$

$= x^0=1\$

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ii) $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}$

$= \frac{\left(1+x^{a-b}\right)+(1+x^{b-a})}{(1+x^{a-b})(1+x^{b-a})}$

$= \frac{2+x^{a-b}+x^{b-a}}{1+x^{a-b}+x^{b-a}+x^{a-b}x^{b-a}}$

$= x^0=1$

Question 6: Simplify:

$\frac{a^{7+2n}.({a^2)}^{3n+2}}{({a^4)}^{2n+3}}$

$= \frac{a^{7+2n+6n+4}}{a^{8n+12}}$

$= a^{7+2n+6n+4-8n-12}$

$= a^{-1}=\frac{1}{a}$

Question 7: Evaluate:

(i) ${\left(\frac{16}{625}\right)}^{1/4}=({\frac{2^4}{5^4})\ }^{\frac{1}{4}}=\ \frac{2}{5}$

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(ii) ${\left(\frac{81}{16}\right)}^{-1/4}=({\frac{3^4}{2^4})\ }^{\frac{-1}{4}}=\ \frac{2}{3}$

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(iii) ${(64)}^{2/3}+\sqrt[3]{125}+3^0+\frac{1}{2^{-5}}+{27}^{-2/3}\times{}{(\frac{25}{9})}^{-1/2}$

$=16+5+1+32+3^{-2}\times{}\frac{3}{5}=54\frac{1}{15}$

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(iv) ${81}^{-1}\times{}3^{-5}\times{}3^9\times{}{64}^{5/6}\times{}({\sqrt[3]{3}}^6)$

$=3^{-4-5+9}4^{5/2}\times{}3^2=288$

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Question 8: Find the value of $x$ when:

i) ${\left(\frac{-3}{11}\right)}^{x+5}\div{}{\left(\frac{-3}{11}\right)}^{-2x+3}={\left(\frac{-3}{11}\right)}^{2x-5}\times{}[({{\frac{-3}{11})}^{-2}]}^{(x+4)}$

${\left(\frac{-3}{11}\right)}^{x+5+2x-3}=\ {\left(\frac{-3}{11}\right)}^{2x-5-2(x+4)}$

$Therefore\ x+5+2x-3=2x-5-2x-8\ or\ x=\ -5$

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ii) ${{{[\{(\frac{2}{5})}^2\}}^4]}^{x+2}=[\{({{{\frac{2}{5})}^{-2}\}}^{\left(x-1\right)}]}^{-3}\$

$2\times{}4\times{}\left(x+2\right)=-2\times{}(x-1)\times{}(-3)$

$8x+16=6x-6\ or\ x=\ -11$

Question 9: Simplify

(i) ${[{\left\{2p^{-1}q^2r\right\}}^3]}^{-2}=\frac{p^6}{64q^{12}r^6}\$

(ii) ${\left(\frac{3p^2qr^{-2}}{2p^{-1}q^3}\right)}^2\div{}{(2p^3r)}^{-1}$

(iii) ${\left(\frac{3p^2qr^{-2}}{2p^{-1}q^3}\right)}^2\times{}{(2p^3r)}^1=\ \frac{9p^9}{2q^4r^3}$