\displaystyle \textbf{Question 1: }\text{Simplify:}
\displaystyle \text{(i) }a^6\times a^8\qquad \text{(ii) }x^5\times x^{-3}\qquad \text{(iii) }z^9\times z^3\times z^{-6}
\displaystyle \text{(iv) }a^2b^3\times a^5b^2\qquad \text{(v) }5x^7\times3x^4\qquad \text{(vi) }p^3q^4\times p^5q^{-5}
\displaystyle \text{(vii) }x^7y^{-5}\times x^{-5}y^3\qquad \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{Answer:}
\displaystyle \text{(i) }a^6\times a^8=a^{6+8}=a^{14}.

\displaystyle \text{(ii) }x^5\times x^{-3}=x^{5-3}=x^2.

\displaystyle \text{(iii) }z^9\times z^3\times z^{-6} =z^{9+3-6}=z^6.

\displaystyle \text{(iv) }a^2b^3\times a^5b^2 =a^{2+5}b^{3+2}=a^7b^5.

\displaystyle \text{(v) }5x^7\times3x^4 =15x^{7+4}=15x^{11}.

\displaystyle \text{(vi) }p^3q^4\times p^5q^{-5} =p^{3+5}q^{4-5}=p^8q^{-1}=\frac{p^8}{q}.

\displaystyle \text{(vii) }x^7y^{-5}\times x^{-5}y^3 =x^{7-5}y^{-5+3}=x^2y^{-2}=\frac{x^2}{y^2}.

\displaystyle \text{(viii) }x^{-2}y^5\times x^0y^{-7} =x^{-2}y^{-2}=\frac{1}{x^2y^2}.

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

\displaystyle \textbf{Question 2: }\text{Simplify:}
\displaystyle \text{(i) }\frac{x^{12}}{x^7}\qquad \text{(ii) }\frac{z^6}{z^{-3}}\qquad \text{(iii) }\frac{m^5}{m^2}\frac{n^3}{n^{-4}}
\displaystyle \text{(iv) }\frac{18x^9}{6x^7}\qquad \text{(v) }\frac{7a^{12}}{56a^{15}}\qquad \text{(vi) }\frac{a^{13}}{a^5}\frac{b^7}{b^{-3}}
\displaystyle \text{(vii) }\frac{7x^{14}}{21x^{-10}}\qquad \text{(viii) }\frac{p^{11}}{p^{11}}\qquad \text{(ix) }\frac{a^7}{a^{-2}}\frac{b^5}{b^3}\frac{c^4}{c^6}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\frac{x^{12}}{x^7} =x^{12-7}=x^5.

\displaystyle \text{(ii) }\frac{z^6}{z^{-3}} =z^{6-(-3)}=z^9.

\displaystyle \text{(iii) }\frac{m^5}{m^2}\frac{n^3}{n^{-4}} =m^{5-2}n^{3-(-4)}=m^3n^7.

\displaystyle \text{(iv) }\frac{18x^9}{6x^7} =3x^{9-7}=3x^2.

\displaystyle \text{(v) }\frac{7a^{12}}{56a^{15}} =\frac{1}{8}a^{-3}=\frac{1}{8a^3}.

\displaystyle \text{(vi) }\frac{a^{13}}{a^5}\frac{b^7}{b^{-3}} =a^{13-5}b^{7-(-3)}=a^8b^{10}.

\displaystyle \text{(vii) }\frac{7x^{14}}{21x^{-10}} =\frac{1}{3}x^{14-(-10)}=\frac{x^{24}}{3}.

\displaystyle \text{(viii) }\frac{p^{11}}{p^{11}} =p^{11-11}=p^0=1.

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

\displaystyle \textbf{Question 3: }\text{Simplify:}
\displaystyle \text{(i) }(a^3)^2\qquad \text{(ii) }(x^2y^{-3})^{-2}\qquad \text{(iii) }(2x^2y)^4
\displaystyle \text{(iv) }(3x^3y^{-3})^2\qquad \text{(v) }(m^2n^{-3})^4
\displaystyle \text{Answer:}
\displaystyle \text{(i) }(a^3)^2=a^{3\times2}=a^6.

\displaystyle \text{(ii) }(x^2y^{-3})^{-2} =x^{-4}y^6=\frac{y^6}{x^4}.

\displaystyle \text{(iii) }(2x^2y)^4 =2^4x^8y^4=16x^8y^4.

\displaystyle \text{(iv) }(3x^3y^{-3})^2 =3^2x^6y^{-6}=\frac{9x^6}{y^6}.

\displaystyle \text{(v) }(m^2n^{-3})^4 =m^8n^{-12}=\frac{m^8}{n^{12}}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Evaluate:}
\displaystyle \text{(i) }36^{1/2}\qquad \text{(ii) }64^{-1/3}\qquad \text{(iii) }27^{2/3}
\displaystyle \text{(iv) }81^{-1/4}\qquad \text{(v) }16^{-3/4}\qquad \text{(vi) }32^{-4/5}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }36^{1/2}=(6^2)^{1/2}=6.

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

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

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

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

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

\displaystyle \textbf{Question 5: }\text{Simplify:}
\displaystyle \text{(i) }(25a^2)^{1/2}\qquad \text{(ii) }(27x^{-3})^{1/3}\qquad \text{(iii) }(64m^{-6}n^3)^{2/3}
\displaystyle \text{(iv) }(81a^4b^8c^{-4})^{1/4}\qquad \text{(v) }(3x^{-3}y^3)^{-2}
\displaystyle \text{(vi) }(6ab^2c^{-3})^{-1}\qquad \text{(vii) }(-3a^{3/4}b^{-1/4})^4
\displaystyle \text{(viii) }(32a^{10}b^{-5})^{1/5}\qquad \text{(ix) }\sqrt[3]{x^{18}y^{-12}z^3}
\displaystyle \text{Answer:}
\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}=3x^{-1}=\frac{3}{x}.

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

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

\displaystyle \text{(v) }(3x^{-3}y^3)^{-2} =3^{-2}x^6y^{-6}=\frac{x^6}{9y^6}.

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

\displaystyle \text{(vii) }(-3a^{3/4}b^{-1/4})^4 =(-3)^4a^3b^{-1}
\displaystyle =81a^3b^{-1}=\frac{81a^3}{b}.

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

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

\displaystyle \textbf{Question 6: }\text{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{Answer:}
\displaystyle \text{(i) LHS}= \frac{x^{m+n}x^{n+l}x^{l+m}}{(x^mx^nx^l)^2}
\displaystyle =\frac{x^{m+n+n+l+l+m}} {(x^{m+n+l})^2}
\displaystyle =\frac{x^{2m+2n+2l}} {x^{2m+2n+2l}}
\displaystyle =1=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}

\displaystyle \text{(ii) LHS}= \sqrt{x^{p-q}}\sqrt{x^{q-r}}\sqrt{x^{r-p}}
\displaystyle =x^{\frac{p-q}{2}} x^{\frac{q-r}{2}}x^{\frac{r-p}{2}}
\displaystyle =x^{\frac{p-q+q-r+r-p}{2}}
\displaystyle =x^0=1=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Show that:}
\displaystyle \text{(i) }(x^{a+b})^{a-b}(x^{b+c})^{b-c} (x^{c+a})^{c-a}=1
\displaystyle \text{(ii) }\left(\frac{x^a}{x^{-b}}\right)^{a-b} \left(\frac{x^b}{x^{-c}}\right)^{b-c} \left(\frac{x^c}{x^{-a}}\right)^{c-a}=1
\displaystyle \text{(iii) }\left(\frac{x^{a+b}}{x^c}\right)^{a-b} \left(\frac{x^{b+c}}{x^a}\right)^{b-c} \left(\frac{x^{c+a}}{x^b}\right)^{c-a}=1
\displaystyle \text{(iv) }\left(\frac{x^{a^2}}{x^{b^2}}\right)^{\frac{1}{a+b}} \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
\displaystyle \text{Answer:}
\displaystyle \text{(i) LHS}=(x^{a+b})^{a-b}(x^{b+c})^{b-c} (x^{c+a})^{c-a}
\displaystyle =x^{(a+b)(a-b)+(b+c)(b-c)+(c+a)(c-a)}
\displaystyle =x^{a^2-b^2+b^2-c^2+c^2-a^2}
\displaystyle =x^0=1=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}

\displaystyle \text{(ii) LHS}= \left(\frac{x^a}{x^{-b}}\right)^{a-b} \left(\frac{x^b}{x^{-c}}\right)^{b-c} \left(\frac{x^c}{x^{-a}}\right)^{c-a}
\displaystyle =(x^{a+b})^{a-b}(x^{b+c})^{b-c} (x^{c+a})^{c-a}
\displaystyle =x^{a^2-b^2+b^2-c^2+c^2-a^2}
\displaystyle =x^0=1=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}

\displaystyle \text{(iii) LHS}= \left(\frac{x^{a+b}}{x^c}\right)^{a-b} \left(\frac{x^{b+c}}{x^a}\right)^{b-c} \left(\frac{x^{c+a}}{x^b}\right)^{c-a}
\displaystyle =(x^{a+b-c})^{a-b}(x^{b+c-a})^{b-c} (x^{c+a-b})^{c-a}
\displaystyle =x^{(a+b-c)(a-b)+(b+c-a)(b-c) +(c+a-b)(c-a)}
\displaystyle =x^{a^2-b^2-ac+bc+b^2-c^2-ab+ac +c^2-a^2-bc+ab}
\displaystyle =x^0=1=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}

\displaystyle \text{(iv) LHS}= \left(\frac{x^{a^2}}{x^{b^2}}\right)^{\frac{1}{a+b}} \left(\frac{x^{b^2}}{x^{c^2}}\right)^{\frac{1}{b+c}} \left(\frac{x^{c^2}}{x^{a^2}}\right)^{\frac{1}{c+a}}
\displaystyle =(x^{a^2-b^2})^{\frac{1}{a+b}} (x^{b^2-c^2})^{\frac{1}{b+c}} (x^{c^2-a^2})^{\frac{1}{c+a}}
\displaystyle =x^{\frac{(a-b)(a+b)}{a+b}} x^{\frac{(b-c)(b+c)}{b+c}} x^{\frac{(c-a)(c+a)}{c+a}}
\displaystyle =x^{a-b}x^{b-c}x^{c-a}
\displaystyle =x^{a-b+b-c+c-a}=x^0=1=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Show that:}
\displaystyle \left(\frac{x^a}{x^b}\right)^{a^2+ab+b^2} \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
\displaystyle \text{Answer:}
\displaystyle \text{LHS}= \left(\frac{x^a}{x^b}\right)^{a^2+ab+b^2} \left(\frac{x^b}{x^c}\right)^{b^2+bc+c^2} \left(\frac{x^c}{x^a}\right)^{c^2+ca+a^2}
\displaystyle =(x^{a-b})^{a^2+ab+b^2} (x^{b-c})^{b^2+bc+c^2} (x^{c-a})^{c^2+ca+a^2}
\displaystyle =x^{(a-b)(a^2+ab+b^2)} x^{(b-c)(b^2+bc+c^2)} x^{(c-a)(c^2+ca+a^2)}
\displaystyle =x^{a^3-b^3}x^{b^3-c^3}x^{c^3-a^3}
\displaystyle =x^{a^3-b^3+b^3-c^3+c^3-a^3}
\displaystyle =x^0=1=\text{RHS}.
\displaystyle \therefore \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Evaluate:}
\displaystyle \text{(i) }\left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}} \left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}} \left(\frac{x^c}{x^a}\right)^{\frac{1}{ca}}
\displaystyle \text{(ii) }\frac{1}{1+x^{a-b}}+ \frac{1}{1+x^{b-a}}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}} \left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}} \left(\frac{x^c}{x^a}\right)^{\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}{1+x^{a-b}} +\frac{1}{1+\frac{1}{x^{a-b}}}
\displaystyle =\frac{1}{1+x^{a-b}} +\frac{x^{a-b}}{1+x^{a-b}}
\displaystyle =\frac{1+x^{a-b}}{1+x^{a-b}}=1.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Simplify: } (ab)^{y-z}(bc)^{z-x}(ca)^{x-z}
\displaystyle \text{Answer:}
\displaystyle (ab)^{y-z}(bc)^{z-x}(ca)^{x-z}
\displaystyle =a^{y-z}b^{y-z}b^{z-x}c^{z-x}c^{x-z}a^{x-z}
\displaystyle =a^{y-z+x-z}b^{y-z+z-x}c^{z-x+x-z}
\displaystyle =a^{x+y-2z}b^{y-x}c^0
\displaystyle =a^{x+y-2z}b^{y-x}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Simplify: } \frac{x^{2n+3}x^{(2n+1)(n+2)}}{(x^3)^{2n+1}x^{n(2n+1)}}
\displaystyle \text{Answer:}
\displaystyle \frac{x^{2n+3}x^{(2n+1)(n+2)}}{(x^3)^{2n+1}x^{n(2n+1)}}
\displaystyle =\frac{x^{2n+3+(2n+1)(n+2)}} {x^{3(2n+1)+n(2n+1)}}
\displaystyle =\frac{x^{2n+3+2n^2+5n+2}} {x^{6n+3+2n^2+n}}
\displaystyle =\frac{x^{2n^2+7n+5}}{x^{2n^2+7n+3}}
\displaystyle =x^{(2n^2+7n+5)-(2n^2+7n+3)}
\displaystyle =x^2.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Simplify: } \frac{a^{7+2n}(a^2)^{3n+2}}{(a^4)^{2n+3}}
\displaystyle \text{Answer:}
\displaystyle \frac{a^{7+2n}(a^2)^{3n+2}}{(a^4)^{2n+3}}
\displaystyle =\frac{a^{7+2n}a^{6n+4}}{a^{8n+12}}
\displaystyle =a^{7+2n+6n+4-8n-12}
\displaystyle =a^{-1}
\displaystyle =\frac{1}{a}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Evaluate:}
\displaystyle \text{(i) }\left(\frac{16}{625}\right)^{1/4}\qquad \text{(ii) }\left(\frac{81}{16}\right)^{-1/4}
\displaystyle \text{(iii) }64^{2/3}+\sqrt[3]{125}+3^0+ \frac{1}{2^{-5}}+27^{-2/3}\left(\frac{25}{9}\right)^{-1/2}
\displaystyle \text{(iv) }81^{-1}\times3^{-5}\times3^9 \times64^{5/6}\times(\sqrt[3]{3})^6
\displaystyle \text{(v) }\sqrt{\frac{y^3}{x}}\times \sqrt{\frac{y}{x}}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\left(\frac{16}{625}\right)^{1/4} =\left(\frac{2^4}{5^4}\right)^{1/4}=\frac{2}{5}.

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

\displaystyle \text{(iii) }64^{2/3}+\sqrt[3]{125}+3^0+ \frac{1}{2^{-5}}+27^{-2/3}\left(\frac{25}{9}\right)^{-1/2}
\displaystyle =16+5+1+32+ 3^{-2}\left(\frac{5^2}{3^2}\right)^{-1/2}
\displaystyle =16+5+1+32+\frac{1}{9}\times\frac{3}{5}
\displaystyle =54+\frac{1}{15}=54\frac{1}{15}.

\displaystyle \text{(iv) }81^{-1}\times3^{-5}\times3^9 \times64^{5/6}\times(\sqrt[3]{3})^6
\displaystyle =3^{-4}\times3^{-5}\times3^9 \times(2^6)^{5/6}\times(3^{1/3})^6
\displaystyle =3^{-4-5+9}\times2^5\times3^2
\displaystyle =1\times32\times9=288.

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

\displaystyle \textbf{Question 15: }\text{Find the value of }x\text{ when:}
\displaystyle \text{(i) }\left(\frac{-3}{11}\right)^{x+5} \div\left(\frac{-3}{11}\right)^{-2x+3} = \left(\frac{-3}{11}\right)^{2x-5} \left[\left(\frac{-3}{11}\right)^{-2}\right]^{x+4}
\displaystyle \text{(ii) } \left[\left\{\left(\frac{2}{5}\right)^2\right\}^4\right]^{x+2} = \left[\left\{\left(\frac{2}{5}\right)^{-2}\right\}^{x-1}\right]^{-3}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\left(\frac{-3}{11}\right)^{x+5} \div\left(\frac{-3}{11}\right)^{-2x+3}
\displaystyle = \left(\frac{-3}{11}\right)^{2x-5} \left[\left(\frac{-3}{11}\right)^{-2}\right]^{x+4}
\displaystyle \left(\frac{-3}{11}\right)^{x+5-(-2x+3)} =\left(\frac{-3}{11}\right)^{2x-5-2(x+4)}
\displaystyle \left(\frac{-3}{11}\right)^{3x+2} =\left(\frac{-3}{11}\right)^{-13}
\displaystyle 3x+2=-13
\displaystyle 3x=-15
\displaystyle \therefore x=-5.

\displaystyle \text{(ii) } \left[\left\{\left(\frac{2}{5}\right)^2\right\}^4\right]^{x+2} =\left[\left\{\left(\frac{2}{5}\right)^{-2}\right\}^{x-1}\right]^{-3}
\displaystyle \left(\frac{2}{5}\right)^{8(x+2)} =\left(\frac{2}{5}\right)^{6(x-1)}
\displaystyle 8(x+2)=6(x-1)
\displaystyle 8x+16=6x-6
\displaystyle 2x=-22
\displaystyle \therefore x=-11.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Simplify:}
\displaystyle \text{(i) }\left[\{2p^{-1}q^2r\}^3\right]^{-2}
\displaystyle \text{(ii) }\left(\frac{3p^2qr^{-2}} {2p^{-1}q^3}\right)^2\div(2p^3r)
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\left[\{2p^{-1}q^2r\}^3\right]^{-2}
\displaystyle =(2p^{-1}q^2r)^{-6}
\displaystyle =2^{-6}p^6q^{-12}r^{-6}
\displaystyle =\frac{p^6}{64q^{12}r^6}.

\displaystyle \text{(ii) }\left(\frac{3p^2qr^{-2}} {2p^{-1}q^3}\right)^2\div(2p^3r)
\displaystyle =\left(\frac{3}{2}p^{2-(-1)} q^{1-3}r^{-2}\right)^2\div(2p^3r)
\displaystyle =\left(\frac{3}{2}p^3q^{-2}r^{-2}\right)^2 \div(2p^3r)
\displaystyle =\frac{9}{4}p^6q^{-4}r^{-4} \times\frac{1}{2p^3r}
\displaystyle =\frac{9}{8}p^3q^{-4}r^{-5}
\displaystyle =\frac{9p^3}{8q^4r^5}.
\displaystyle \\

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