Question 1: Simplify:

$\displaystyle \text{i) } a^6 \times a^8$          $\displaystyle \text{ii) } x^5 \times x^{-3}$         $\displaystyle \text{iii) } z^9 \times z^3 \times z^{-6}$

$\displaystyle \text{iv) } a^2b^3 \times a^5b^2$         $\displaystyle \text{v) } 5x^7 \times 3x^4$         $\displaystyle \text{vi) } p^3q^4 \times p^5q^{-5}$

$\displaystyle \text{vii) } x^7y^{-5} \times x^{-5}y^3$         $\displaystyle \text{viii) } x^{-2}y^5 \times x^0y^{-7}$         $\displaystyle \text{ix) } x^6y^4z^{-2} \times x^{-3}y^{-5}z^{-1} \times x^2z^4$

$\displaystyle \text{i) } a^6 \times a^8 = a^{14}$ $\displaystyle \text{ii) } x^5 \times x^{-3} = x^2$

$\displaystyle \text{iii) } z^9 \times z^3 \times z^{-6} = z^6$ $\displaystyle \text{iv) } a^2b^3 \times a^5b^2 = a^7b^5$

$\displaystyle \text{v) } 5x^7 \times 3x^4 = 15x^{11}$ $\displaystyle \text{vi) } p^3q^4 \times p^5q^{-5} = p^8q^{-1}$

$\displaystyle \text{vii) } x^7y^{-5} \times x^{-5}y^3 = x^2y^{-2}$ $\displaystyle \text{viii) } x^{-2}y^5 \times x^0y^{-7} = x^{-2}y^{-2}$

$\displaystyle \text{ix) } x^6y^4z^{-2} \times x^{-3}y^{-5}z^{-1} \times x^2z^4 = x^5y^{-1}z^1$

$\displaystyle \\$

Question 2: Simplify:

$\displaystyle \text{i) } \frac{x^{12}}{x^7}$         $\displaystyle \text{ii) } \frac{z^6}{z^{-3}}$         $\displaystyle \text{iii) } \frac{m^5}{m^2}\frac{n^3}{n^{-4}}$         $\displaystyle \text{iv) } \frac{18x^9}{6x^7}$         $\displaystyle \text{v) } \frac{7a^{12}}{56a^{15}}$         $\displaystyle \text{vi) } \frac{a^{13}}{a^5}\frac{b^7}{b^{-3}}$

$\displaystyle \text{vii) } \frac{7x^{14}}{21x^{-10}}$         $\displaystyle \text{viii) } \frac{p^{11}}{p^{11}}$         $\displaystyle \text{ix) } \frac{a^7}{a^{-2}}\frac{b^5}{b^3}\frac{c^4}{c^6}$

$\displaystyle \text{i) } \frac{x^{12}}{x^7} = x^5$ $\displaystyle \text{ii) } \frac{z^6}{z^{-3}} = z^9$

$\displaystyle \text{iii) } \frac{m^5}{m^2}\frac{n^3}{n^{-4}} = m^3n^7$ $\displaystyle \text{iv) } \frac{18x^9}{6x^7} = 3x^2$

$\displaystyle \text{v) } \frac{7a^{12}}{56a^{15}} = \frac{a^{-3}}{8}$ $\displaystyle \text{vi) } \frac{a^{13}}{a^5}\frac{b^7}{b^{-3}} = a^8b^{10}$

$\displaystyle \text{vii) } \frac{7x^{14}}{21x^{-10}} = \frac{x^{24}}{3}$ $\displaystyle \text{viii) } \frac{p^{11}}{p^{11}} = p^0 = 1$

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

$\displaystyle \\$

Question 3: Simplify:

$\displaystyle \text{i) } {(a^3)}^2$         $\displaystyle \text{ii) } {(x^2y^{-3})}^{-2}$         $\displaystyle \text{iii) } {(2x^2y)}^4$         $\displaystyle \text{iv) } {(3x^3y^{-3})}^2$         $\displaystyle \text{v) } {(m^2n^{-3})}^4$

$\displaystyle \text{i) } {(a^3)}^2 = a^6$ $\displaystyle \text{ii) } {(x^2y^{-3})}^{-2} = x^{-4}y^6$ $\displaystyle \text{iii) } {(2x^2y)}^4 = 16x^8y^4$

$\displaystyle \text{iv) } {(3x^3y^{-3})}^2 = 9x^6y^{-6}$ $\displaystyle \text{v) } {(m^2n^{-3})}^4 = m^8n^{-12}$

$\displaystyle \\$

Question 4: Evaluate

$\displaystyle \text{i) } {(36)}^{1/2}$         $\displaystyle \text{ii) } {(64)}^{-1/3}$         $\displaystyle \text{iii) } {(27)}^{2/3}$         $\displaystyle \text{iv) } {(81)}^{-1/4}$         $\displaystyle \text{v) } {(16)}^{-3/4}$         $\displaystyle \text{vi) } {(32)}^{-4/5}$

$\displaystyle \text{i) } {(36)}^{1/2} = {(6^2)}^{1/2} = 6$

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

$\displaystyle \text{iii) } {(27)}^{2/3} = {(3^3)}^{2/3} = 9$

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

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

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

$\displaystyle \\$

Question 5: Simplify

$\displaystyle \text{i) } {(25a^2)}^{1/2}$         $\displaystyle \text{ii) } {(27x^{-3})}^{1/3}$         $\displaystyle \text{iii) } {(64m^{-6}n^3)}^{2/3}$         $\displaystyle \text{iv) } {(81a^4b^8c^{-4})}^{1/4}$

$\displaystyle \text{v) } {(3x^{-3}y^3)}^{-2}$         $\displaystyle \text{vi) } {(6ab^2c^{-3})}^{-1}$         $\displaystyle \text{vii) } {(-3a^{3/4}b^{-1/4})}^4$         $\displaystyle \text{viii) } {(32a^{10}b^{-5})}^{1/5}$

$\displaystyle \text{ix) } \sqrt[3]{x^{18}y^{-12}z^3}$

$\displaystyle \text{i) } {(25a^2)}^{1/2} = {(5^2a^2)}^{1/2} = 5a$

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 6: Show that:

$\displaystyle \text{i) } \frac{x^{m+n}x^{n+l}x^{l+m}}{{(x^mx^nx^l)}^2} = 1$         $\displaystyle \text{ii) } \sqrt{x^{p-q}}\sqrt{x^{q-r}}\sqrt{x^{r-p}} = 1$

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

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

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

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

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

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

$\displaystyle \Rightarrow { (\frac{x^p}{x^q})}^r \times {(\frac{x^q}{x^r})}^p \times {(\frac{x^r}{x^p})}^q = 1$

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

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

$\displaystyle \Rightarrow x^0 = 1$ Hence proved.

$\displaystyle \\$

Question 8: Show that:

$\displaystyle \text{i) } {{(x}^{a+b})}^{ (a-b )}({x^{b+c})}^{ (b-c )}{{(x}^{c+a})}^{ (c-a )} = 1$

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

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

$\displaystyle \text{iv) } { (\frac{x^{a^2}}{x^{b^2}} )}^{\frac{1}{a+b}}{ \times (\frac{x^{b^2}}{x^{c^2}} )}^{\frac{1}{b+c}} \times { (\frac{x^{c^2}}{x^{a^2}} )}^{\frac{1}{c+a}} = 1$

$\displaystyle \text{i) } {{(x}^{a+b})}^{ (a-b )}({x^{b+c})}^{ (b-c )}{{(x}^{c+a})}^{ (c-a )} = 1$

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

$\displaystyle \Rightarrow x^0 = 1$ Hence proved.

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

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

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

$\displaystyle \Rightarrow x^0 = 1$ Hence proved.

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

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

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

$\displaystyle \Rightarrow x^0 = 1$ Hence proved.

$\displaystyle \text{iv) } { (\frac{x^{a^2}}{x^{b^2}} )}^{\frac{1}{a+b}}{ \times (\frac{x^{b^2}}{x^{c^2}} )}^{\frac{1}{b+c}} \times { (\frac{x^{c^2}}{x^{a^2}} )}^{\frac{1}{c+a}} = 1$

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

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

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

$\displaystyle \Rightarrow x^0 = 1$ Hence proved.

$\displaystyle \\$

Question 9: Show that:

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

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

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

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

$\displaystyle x^0 = 1$ Hence proved.

$\displaystyle \\$

Question 10: Evaluate:

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

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

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

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

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

$\displaystyle = x^0 = 1$

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

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

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

$\displaystyle = x^0 = 1$

$\displaystyle \\$

Question 11: Simplify: $\displaystyle (ab)^{y-z} . (bc)^{z-x} . (ca)^{x-z}$

$\displaystyle (ab)^{y-z} \ . \ (bc)^{z-x} \ . \ (ca)^{x-z}$

$\displaystyle = a^{y-z+x-y} \ . \ b^{y-z+z-x} \ . \ c^{z-x+x-y}$

$\displaystyle = a^{x-z} b^{y-x}c^{z-y}$

$\displaystyle \\$

Question 12: $\displaystyle \text{Simplify: } \frac{x^{2n+3} . x^{(2n+1)(n+2)}}{(x^3)^{2n+1} . x^{n(2n+1)}}$

$\displaystyle \frac{x^{2n+3} . x^{(2n+1)(n+2)}}{(x^3)^{2n+1} . x^{n(2n+1)}} = \frac{x^{2n+3+2n^2+4n+n+2} }{ x^{6n+3+2n^2+n} } = \frac{ x^{2n^2+7n + 5}}{x^{2n^2+7n + 3}} =x^2$

$\displaystyle \\$

Question 13: Simplify: $\displaystyle \frac{a^{7+2n}.({a^2)}^{3n+2}}{({a^4)}^{2n+3}}$

$\displaystyle \frac{a^{7+2n}.({a^2)}^{3n+2}}{({a^4)}^{2n+3}} = \frac{a^{7+2n+6n+4}}{a^{8n+12}} = a^{7+2n+6n+4-8n-12} = a^{-1} = \frac{1}{a}$

$\displaystyle \\$

Question 14: Evaluate:

$\displaystyle \text{i) } { (\frac{16}{625} )}^{1/4}$         $\displaystyle \text{ii) } { (\frac{81}{16} )}^{-1/4}$         $\displaystyle \text{iii) } {(64)}^{2/3}+\sqrt[3]{125}+3^0+\frac{1}{2^{-5}}+{27}^{-2/3} \times {(\frac{25}{9})}^{-1/2}$

$\displaystyle \text{iv) } {81}^{-1} \times 3^{-5} \times 3^9 \times {64}^{5/6} \times ({\sqrt[3]{3}}^6)$         $\displaystyle \text{v) } \sqrt{\frac{y^3}{x}} \times \sqrt{\frac{y}{x}}$

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

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

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

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

$\displaystyle \text{iv) } {81}^{-1} \times 3^{-5} \times 3^9 \times {64}^{5/6} \times ({\sqrt[3]{3}}^6) = 3^{-4-5+9}4^{5/2} \times 3^2 = 288$

$\displaystyle \text{v) } \sqrt{\frac{y^3}{x}} \times \sqrt{\frac{y}{x}} = \frac{y^{\frac{3}{2}}}{x^{\frac{1}{2}}} . \frac{y^{\frac{1}{2}}}{x^{\frac{1}{2}}} = \frac{y^2}{x}$

$\displaystyle \\$

Question 15: Find the value of $\displaystyle x$ when:

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

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

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

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

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

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

$\displaystyle 2 \times 4 \times ( x+2 ) =-2 \times (x-1) \times (-3)$

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

$\displaystyle \\$

Question 16: Simplify

$\displaystyle \text{i) } { \Big[ {\left\{2p^{-1}q^2r\right\}}^3 \Big] }^{-2}$

$\displaystyle \text{ii) } \Big( \frac{3p^2 qr^{-2} }{2p^{-1} q^3} \Big)^2 \div {(2p^3r)}^1$

$\displaystyle \text{i) } { \Big[ {\left\{2p^{-1}q^2r\right\}}^3 \Big] }^{-2}= \frac{p^6}{64q^{12}r^6}$
$\displaystyle \text{ii) } \Big( \frac{3p^2 qr^{-2} }{2p^{-1} q^3} \Big)^2 \div {(2p^3r)}^1 = \Big( \frac{3p^2 qr^{-2} }{2p^{-1} q^3} \Big)^2 \times {(2p^3r)}^1 = \frac{9p^9}{2q^4r^3}$