Class 9: Indices – Exercise 8

$\displaystyle \textbf{Exercise 8(A)}$

$\displaystyle \textbf{Question 1:}$
$\displaystyle \text{(i) }125^{\frac13}\qquad \text{(ii) }8^{\frac23}\qquad \text{(iii) }\left(\frac15\right)^{-2}$
$\displaystyle \text{(iv) }16^{-\frac34}\qquad \text{(v) }32^{-\frac45}\qquad \text{(vi) }\left(\frac8{125}\right)^{-\frac13}$
$\displaystyle \text{(vii) }(-27)^{\frac23}\qquad \text{(viii) }(0.001)^{-\frac13}\qquad \text{(ix) }(0.027)^{-\frac23}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }125^{\frac13}=(5\times5\times5)^{\frac13}$
$\displaystyle =\left(5^3\right)^{\frac13}$
$\displaystyle =5^{3\times\frac13}$
$\displaystyle =5$
$\displaystyle \\$

$\displaystyle \text{(ii) }8^{\frac23}=(2\times2\times2)^{\frac23}$
$\displaystyle =\left(2^3\right)^{\frac23}$
$\displaystyle =2^{3\times\frac23}$
$\displaystyle =2^2$
$\displaystyle =4$
$\displaystyle \\$

$\displaystyle \text{(iii) }\left(\frac15\right)^{-2}=\left(\frac51\right)^2$
$\displaystyle =5^2$
$\displaystyle =25$
$\displaystyle \\$

$\displaystyle \text{(iv) }16^{-\frac34}=(2\times2\times2\times2)^{-\frac34}$
$\displaystyle =\left(2^4\right)^{-\frac34}$
$\displaystyle =2^{4\times\left(-\frac34\right)}$
$\displaystyle =2^{-3}$
$\displaystyle =\frac{1}{2^3}$
$\displaystyle =\frac18$
$\displaystyle \\$

$\displaystyle \text{(v) }32^{-\frac45}=(2\times2\times2\times2\times2)^{-\frac45}$
$\displaystyle =\left(2^5\right)^{-\frac45}$
$\displaystyle =2^{5\times\left(-\frac45\right)}$
$\displaystyle =2^{-4}$
$\displaystyle =\frac{1}{2^4}$
$\displaystyle =\frac1{16}$
$\displaystyle \\$

$\displaystyle \text{(vi) }\left(\frac8{125}\right)^{-\frac13}$
$\displaystyle =\left(\frac{2\times2\times2}{5\times5\times5}\right)^{-\frac13}$
$\displaystyle =\left[\left(\frac25\right)^3\right]^{-\frac13}$
$\displaystyle =\left(\frac25\right)^{3\times\left(-\frac13\right)}$
$\displaystyle =\left(\frac25\right)^{-1}$
$\displaystyle =\frac52$
$\displaystyle \\$

$\displaystyle \text{(vii) }(-27)^{\frac23}=\left[(-3)\times(-3)\times(-3)\right]^{\frac23}$
$\displaystyle =\left((-3)^3\right)^{\frac23}$
$\displaystyle =(-3)^{3\times\frac23}$
$\displaystyle =(-3)^2$
$\displaystyle =9$
$\displaystyle \\$

$\displaystyle \text{(viii) }(0.001)^{-\frac13}$
$\displaystyle =(0.1\times0.1\times0.1)^{-\frac13}$
$\displaystyle =\left((0.1)^3\right)^{-\frac13}$
$\displaystyle =(0.1)^{-1}$
$\displaystyle =\frac{1}{0.1}$
$\displaystyle =10$
$\displaystyle \\$

$\displaystyle \text{(ix) }(0.027)^{-\frac23}$
$\displaystyle =(0.3\times0.3\times0.3)^{-\frac23}$
$\displaystyle =\left((0.3)^3\right)^{-\frac23}$
$\displaystyle =(0.3)^{-2}$
$\displaystyle =\frac{1}{(0.3)^2}$
$\displaystyle =\frac{1}{0.09}$
$\displaystyle =\frac{100}{9}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Evaluate the following:}$
$\displaystyle \text{(i) }\left(\frac14\right)^{-2}-3\times8^{\frac23}\times5^0+\left(\frac9{16}\right)^{-\frac12}$
$\displaystyle \text{(ii) }\sqrt{\frac14}+(0.01)^{-\frac12}-(27)^{\frac23}\times3^0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(\frac14\right)^{-2}-3\times8^{\frac23}\times5^0+\left(\frac9{16}\right)^{-\frac12}$
$\displaystyle =\left(\frac12\times\frac12\right)^{-2}-3\times(2^3)^{\frac23}\times5^0+\left(\frac34\times\frac34\right)^{-\frac12}$
$\displaystyle =\left(\frac12\right)^{2\times(-2)}-3\times2^{3\times\frac23}\times5^0+\left(\frac34\right)^{2\times\left(-\frac12\right)}$
$\displaystyle =\left(\frac12\right)^{-4}-3\times2^2\times1+\left(\frac34\right)^{-1}$
$\displaystyle =2^4-3\times4\times1+\frac43$
$\displaystyle =16-12+\frac43$
$\displaystyle =4+\frac43$
$\displaystyle =\frac{12+4}{3}$
$\displaystyle =\frac{16}{3}$
$\displaystyle =5\frac13$

$\displaystyle \text{(ii) }\sqrt{\frac14}+(0.01)^{-\frac12}-(27)^{\frac23}\times3^0$
$\displaystyle =\left[\left(\frac12\right)^2\right]^{\frac12}+(0.1\times0.1)^{-\frac12}-(3\times3\times3)^{\frac23}\times3^0$
$\displaystyle =\left(\frac12\right)^{2\times\frac12}+(0.1)^{2\times\left(-\frac12\right)}-(3^3)^{\frac23}\times3^0$
$\displaystyle =\left(\frac12\right)^1+(0.1)^{-1}-3^2\times3^0$
$\displaystyle =\frac12+\frac1{0.1}-9\times1$
$\displaystyle =\frac12+\frac{10}{1}-9$
$\displaystyle =10\frac12-9$
$\displaystyle =1\frac12$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Evaluate the following:}$
$\displaystyle \text{(i) }\left(\frac{81}{16}\right)^{-\frac34}\times\left[\left(\frac{25}{9}\right)^{-\frac32}\div\left(\frac52\right)^{-3}\right]$
$\displaystyle \text{(ii) }\left[(64)^{\frac23}\times2^{-2}\div7^0\right]^{-\frac12}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(\frac{81}{16}\right)^{-\frac34}\times\left[\left(\frac{25}{9}\right)^{-\frac32}\div\left(\frac52\right)^{-3}\right]$
$\displaystyle =\left(\frac{3\times3\times3\times3}{2\times2\times2\times2}\right)^{-\frac34}$
$\displaystyle \qquad\times\left[\left(\frac{5\times5}{3\times3}\right)^{-\frac32}\div\left(\frac52\right)^{-3}\right]$
$\displaystyle =\left[\left(\frac32\right)^4\right]^{-\frac34}\times\left[\left(\frac53\right)^2\right]^{-\frac32}\div\left(\frac52\right)^{-3}$
$\displaystyle =\left(\frac32\right)^{-3}\times\left[\left(\frac53\right)^{-3}\div\left(\frac52\right)^{-3}\right]$
$\displaystyle =\left(\frac23\right)^3\times\left[\left(\frac35\right)^3\div\left(\frac25\right)^3\right]$
$\displaystyle =\left(\frac23\right)^3\times\left[\left(\frac35\right)^3\times\left(\frac52\right)^3\right]$
$\displaystyle =\frac{8}{27}\times\left(\frac32\right)^3$
$\displaystyle =\frac{8}{27}\times\frac{27}{8}$
$\displaystyle =1$

$\displaystyle \text{(ii) }\left[(64)^{\frac23}\times2^{-2}\div7^0\right]^{-\frac12}$
$\displaystyle =\left[\left(4\times4\times4\right)^{\frac23}\times\frac1{2^2}\div1\right]^{-\frac12}$
$\displaystyle =\left[\left(4^3\right)^{\frac23}\times\frac14\right]^{-\frac12}$
$\displaystyle =\left[4^{3\times\frac23}\times\frac14\right]^{-\frac12}$
$\displaystyle =\left[4^2\times\frac14\right]^{-\frac12}$
$\displaystyle =\left[16\times\frac14\right]^{-\frac12}$
$\displaystyle =4^{-\frac12}$
$\displaystyle =\frac{1}{4^{\frac12}}$
$\displaystyle =\frac12$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Evaluate the following:}$
$\displaystyle \text{(i) }(81)^{\frac34}-\left(\frac1{32}\right)^{-\frac25}+(8)^{\frac13}\left(\frac12\right)^{-1}(2)^0$
$\displaystyle \text{(ii) }\left(\frac{16}{81}\right)^{-\frac34}\times\left(\frac{49}{9}\right)^{\frac32}\div\left(\frac{343}{216}\right)^{\frac23}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(81)^{\frac34}-\left(\frac1{32}\right)^{-\frac25}+(8)^{\frac13}\left(\frac12\right)^{-1}(2)^0$
$\displaystyle =(3\times3\times3\times3)^{\frac34}-\left(\frac1{2\times2\times2\times2\times2}\right)^{-\frac25}$
$\displaystyle \qquad+(2\times2\times2)^{\frac13}\times\left(\frac12\right)^{-1}\times2^0$
$\displaystyle =\left(3^4\right)^{\frac34}-\left(\frac1{2^5}\right)^{-\frac25}+\left(2^3\right)^{\frac13}\times\left(\frac12\right)^{-1}\times1$
$\displaystyle =3^{4\times\frac34}-\left(2^{-5}\right)^{-\frac25}+2^{3\times\frac13}\times2^1$
$\displaystyle =3^3-2^2+2^1\times2^1$
$\displaystyle =27-4+4$
$\displaystyle =27$

$\displaystyle \text{(ii) }\left(\frac{16}{81}\right)^{-\frac34}\times\left(\frac{49}{9}\right)^{\frac32}\div\left(\frac{343}{216}\right)^{\frac23}$
$\displaystyle =\left(\frac{2\times2\times2\times2}{3\times3\times3\times3}\right)^{-\frac34}\times\left(\frac{7\times7}{3\times3}\right)^{\frac32}$
$\displaystyle \qquad\div\left(\frac{7\times7\times7}{6\times6\times6}\right)^{\frac23}$
$\displaystyle =\left[\left(\frac23\right)^4\right]^{-\frac34}\times\left[\left(\frac73\right)^2\right]^{\frac32}\div\left[\left(\frac76\right)^3\right]^{\frac23}$
$\displaystyle =\left(\frac23\right)^{-3}\times\left(\frac73\right)^3\div\left(\frac76\right)^2$
$\displaystyle =\left(\frac32\right)^3\times\frac{7\times7\times7}{3\times3\times3}\times\frac{6\times6}{7\times7}$
$\displaystyle =\frac{3\times3\times3}{2\times2\times2}\times\frac{343}{27}\times\frac{36}{49}$
$\displaystyle =\frac{27}{8}\times\frac{343}{27}\times\frac{36}{49}$
$\displaystyle =\frac{63}{2}$
$\displaystyle =31\frac12$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Evaluate the following:}$
$\displaystyle \text{(i) }\left(\frac{64}{125}\right)^{-\frac23}\div\frac{1}{\left(\frac{256}{625}\right)^{\frac14}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^0$
$\displaystyle \text{(ii) }\frac{(32)^{\frac25}\times(4)^{-\frac12}\times(8)^{\frac13}}{(2)^{-2}+(64)^{-\frac13}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\left(\frac{64}{125}\right)^{-\frac23}\div\frac{1}{\left(\frac{256}{625}\right)^{\frac14}}+\left(\frac{\sqrt{25}}{\sqrt[3]{64}}\right)^0$
$\displaystyle =\left(\frac{4\times4\times4}{5\times5\times5}\right)^{-\frac23}\div\frac{1}{\left(\frac{4\times4\times4\times4}{5\times5\times5\times5}\right)^{\frac14}}+\left(\frac{\sqrt{5^2}}{\sqrt[3]{4^3}}\right)^0$
$\displaystyle =\left[\left(\frac45\right)^3\right]^{-\frac23}\div\frac{1}{\left[\left(\frac45\right)^4\right]^{\frac14}}+\left(\frac54\right)^0$
$\displaystyle =\left(\frac45\right)^{-2}\div\frac{1}{\left(\frac45\right)^1}+1$
$\displaystyle =\left(\frac54\right)^2\times\frac45+1$
$\displaystyle =\frac{25}{16}\times\frac45+1$
$\displaystyle =\frac54+1$
$\displaystyle =\frac94$
$\displaystyle =2\frac14$
$\displaystyle \\$

$\displaystyle \text{(ii) }\frac{(32)^{\frac25}\times(4)^{-\frac12}\times(8)^{\frac13}}{(2)^{-2}+(64)^{-\frac13}}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{(32)^{\frac25}\times(4)^{-\frac12}\times(8)^{\frac13}}{(2)^{-2}+(64)^{-\frac13}}$
$\displaystyle =\frac{(2\times2\times2\times2\times2)^{\frac25}\times(2\times2)^{-\frac12}\times(2\times2\times2)^{\frac13}}{(2)^{-2}+(4\times4\times4)^{-\frac13}}$
$\displaystyle =\frac{(2^5)^{\frac25}\times(2^2)^{-\frac12}\times(2^3)^{\frac13}}{(2)^{-2}+(4^3)^{-\frac13}}$
$\displaystyle =\frac{(2)^{5\times\frac25}\times(2)^{2\times\left(-\frac12\right)}\times(2)^{3\times\frac13}}{(2)^{-2}+(4)^{3\times\left(-\frac13\right)}}$
$\displaystyle =\frac{2^2\times2^{-1}\times2^1}{2^{-2}+4^{-1}}$
$\displaystyle =\frac{2^{2-1+1}}{2^{-2}+2^{-2}}$
$\displaystyle =\frac{2^2}{2^{-2}+2^{-2}}$
$\displaystyle =\frac{2\times2}{\frac14+\frac14}$
$\displaystyle =\frac{4}{\frac12}$
$\displaystyle =8$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Evaluate the following:}$
$\displaystyle \text{(i) }(27)^{\frac43}+(32)^{0.8}+(0.8)^{-1}+(0.8)^0$
$\displaystyle \text{(ii) }\left[\frac{(27)^{-3}}{(9)^{-3}}\right]^{\frac13}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(27)^{\frac43}+(32)^{0.8}+(0.8)^{-1}+(0.8)^0$
$\displaystyle =(3\times3\times3)^{\frac43}+(2\times2\times2\times2\times2)^{\frac45}+\frac1{0.8}+1$
$\displaystyle =3^{3\times\frac43}+2^{5\times\frac45}+\frac54+1$
$\displaystyle =3^4+2^4+\frac54+1$
$\displaystyle =81+16+\frac54+1$
$\displaystyle =98+\frac54$
$\displaystyle =99\frac14$

$\displaystyle \text{(ii) }\left[\frac{(27)^{-3}}{(9)^{-3}}\right]^{\frac13}$
$\displaystyle =\left[\left(\frac{27}{9}\right)^{-3}\right]^{\frac13}$
$\displaystyle =(3^{-3})^{\frac13}$
$\displaystyle =3^{-1}$
$\displaystyle =\frac13$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Evaluate the following:}$
$\displaystyle \text{(i) }(\sqrt{32}-\sqrt5)^{\frac13}(\sqrt{32}+\sqrt5)^{\frac13}$
$\displaystyle \text{(ii) }9^{\frac52}-3(4)^0-\left(\frac1{81}\right)^{-\frac12}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(\sqrt{32}-\sqrt5)^{\frac13}(\sqrt{32}+\sqrt5)^{\frac13}$
$\displaystyle =\left[(\sqrt{32}-\sqrt5)(\sqrt{32}+\sqrt5)\right]^{\frac13}$
$\displaystyle =\left[(\sqrt{32})^2-(\sqrt5)^2\right]^{\frac13}$
$\displaystyle =(32-5)^{\frac13}$
$\displaystyle =(27)^{\frac13}$
$\displaystyle =(3^3)^{\frac13}$
$\displaystyle =3$

$\displaystyle \text{(ii) }9^{\frac52}-3(4)^0-\left(\frac1{81}\right)^{-\frac12}$
$\displaystyle =(3\times3)^{\frac52}-3\times1-\left(\frac1{9\times9}\right)^{-\frac12}$
$\displaystyle =(3^2)^{\frac52}-3-\left(\frac19\right)^{-1}$
$\displaystyle =3^{2\times\frac52}-3-9$
$\displaystyle =3^5-12$
$\displaystyle =243-12$
$\displaystyle =231$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Simplify: }\frac{3^n\times9^{n+1}}{3^{n-1}\times9^{n-1}}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{3^n\times9^{n+1}}{3^{n-1}\times9^{n-1}}$
$\displaystyle =\frac{3^n\times\left(3^2\right)^{n+1}}{3^{n-1}\times\left(3^2\right)^{n-1}}$
$\displaystyle =\frac{3^n\times3^{2n+2}}{3^{n-1}\times3^{2n-2}}$
$\displaystyle =\frac{3^{3n+2}}{3^{3n-3}}$
$\displaystyle =3^{3n+2-(3n-3)}$
$\displaystyle =3^5$
$\displaystyle =243$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Simplify: }\frac{(27)^{\frac{2n}{3}}\times(8)^{-\frac n6}}{(18)^{-\frac n2}}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{(27)^{\frac{2n}{3}}\times(8)^{-\frac n6}}{(18)^{-\frac n2}}$
$\displaystyle =\frac{\left(3^3\right)^{\frac{2n}{3}}\times\left(2^3\right)^{-\frac n6}}{\left(2\times3^2\right)^{-\frac n2}}$
$\displaystyle =\frac{3^{3\times\frac{2n}{3}}\times2^{3\times\left(-\frac n6\right)}}{2^{-\frac n2}\times3^{-n}}$
$\displaystyle =\frac{3^{2n}\times2^{-\frac n2}}{2^{-\frac n2}\times3^{-n}}$
$\displaystyle =3^{2n}\times3^n$
$\displaystyle =3^{3n}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Simplify: }\frac{5^{2(n+6)}\times(25)^{-7+2n}}{(125)^{2n}}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{5^{2(n+6)}\times(25)^{-7+2n}}{(125)^{2n}}$
$\displaystyle =\frac{5^{2n+12}\times\left(5^2\right)^{-7+2n}}{\left(5^3\right)^{2n}}$
$\displaystyle =\frac{5^{2n+12}\times5^{-14+4n}}{5^{6n}}$
$\displaystyle =\frac{5^{6n-2}}{5^{6n}}$
$\displaystyle =5^{-2}$
$\displaystyle =\frac{1}{5^2}$
$\displaystyle =\frac1{25}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Simplify: }\frac{5^{n+3}-16\times5^{n+1}}{12\times5^n-2\times5^{n+1}}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{5^{n+3}-16\times5^{n+1}}{12\times5^n-2\times5^{n+1}}$
$\displaystyle =\frac{5^n\left(5^3-16\times5\right)}{5^n\left(12-2\times5\right)}$
$\displaystyle =\frac{125-80}{12-10}$
$\displaystyle =\frac{45}{2}$
$\displaystyle =22.5$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Simplify: }\frac{3\times(27)^{n+1}+9\times3^{3n-1}}{8\times3^{3n}-5\times(27)^n}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{3\times(27)^{n+1}+9\times3^{3n-1}}{8\times3^{3n}-5\times(27)^n}$
$\displaystyle =\frac{3\times(3^3)^{n+1}+9\times3^{3n-1}}{8\times3^{3n}-5\times(3^3)^n}$
$\displaystyle =\frac{3\times3^{3n+3}+9\times3^{3n-1}}{8\times3^{3n}-5\times3^{3n}}$
$\displaystyle =\frac{3^{3n}\left(3\times3^3+9\times3^{-1}\right)}{3^{3n}(8-5)}$
$\displaystyle =\frac{3\times27+9\times\frac13}{3}$
$\displaystyle =\frac{81+3}{3}$
$\displaystyle =\frac{84}{3}$
$\displaystyle =28$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Simplify: }\frac{5\times(25)^{n+1}-25\times5^{2n}}{5\times5^{2n+3}-(25)^{n+1}}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{5\times(25)^{n+1}-25\times5^{2n}}{5\times5^{2n+3}-(25)^{n+1}}$
$\displaystyle =\frac{5\times(5^2)^{n+1}-25\times5^{2n}}{5\times5^{2n+3}-(5^2)^{n+1}}$
$\displaystyle =\frac{5\times5^{2n+2}-25\times5^{2n}}{5\times5^{2n+3}-5^{2n+2}}$
$\displaystyle =\frac{5^{2n}(5\times5^2-25)}{5^{2n}(5\times5^3-5^2)}$
$\displaystyle =\frac{5\times25-25}{5\times125-25}$
$\displaystyle =\frac{125-25}{625-25}$
$\displaystyle =\frac{100}{600}$
$\displaystyle =\frac16$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Simplify: }\frac{7^{2n+3}-(49)^{n+2}}{\left[(343)^{n+1}\right]^{\frac23}}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{7^{2n+3}-(49)^{n+2}}{\left[(343)^{n+1}\right]^{\frac23}}$
$\displaystyle =\frac{7^{2n+3}-(7^2)^{n+2}}{\left[(7^3)^{n+1}\right]^{\frac23}}$
$\displaystyle =\frac{7^{2n+3}-7^{2n+4}}{7^{3(n+1)\times\frac23}}$
$\displaystyle =\frac{7^{2n}(7^3-7^4)}{7^{2n+2}}$
$\displaystyle =\frac{7^3-7^4}{7^2}$
$\displaystyle =\frac{7^3(1-7)}{7^2}$
$\displaystyle =7(1-7)$
$\displaystyle =7(-6)$
$\displaystyle =-42$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Simplify: }\left(x^{\frac13}-x^{-\frac13}\right)\left(x^{\frac23}+1+x^{-\frac23}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x^{\frac13}=a,\text{ so that }x^{-\frac13}=\frac1a.$
$\displaystyle \therefore \left(x^{\frac13}-x^{-\frac13}\right)\left(x^{\frac23}+1+x^{-\frac23}\right)$
$\displaystyle =\left(a-\frac1a\right)\left(a^2+1+\frac1{a^2}\right)$
$\displaystyle =a^3-\frac1{a^3}\qquad\left[\because\ (a-b)(a^2+ab+b^2)=a^3-b^3,\ b=\frac1a\right]$
$\displaystyle =\left(x^{\frac13}\right)^3-\left(x^{-\frac13}\right)^3$
$\displaystyle =x-\frac1x$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{Simplify:}$
$\displaystyle \text{(i) }a^7\times a^4\times a^{-6}\times a^0$
$\displaystyle \text{(ii) }a^{\frac43}\div a^{-\frac23}$
$\displaystyle \text{(iii) }\left(a^{-1}+b^{-1}\right)\div\left(a^{-2}-b^{-2}\right)$
$\displaystyle \text{(iv) }\left(a^{-1}+b^{-1}\right)\div(ab)^{-1}$
$\displaystyle \text{(v) }\left(a^{-1}\times b^{-1}\right)\div\left(a^{-1}+b^{-1}\right)$
$\displaystyle \text{(vi) }(a+b)^{-1}\times\left(a^{-1}+b^{-1}\right)$
$\displaystyle \text{(vii) }\frac{a+b+c}{a^{-1}b^{-1}+b^{-1}c^{-1}+c^{-1}a^{-1}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }a^7\times a^4\times a^{-6}\times a^0$
$\displaystyle =a^{7+4-6+0}$
$\displaystyle =a^5$

$\displaystyle \text{(ii) }a^{\frac43}\div a^{-\frac23}$
$\displaystyle =a^{\frac43-\left(-\frac23\right)}$
$\displaystyle =a^{\frac43+\frac23}$
$\displaystyle =a^{\frac63}$
$\displaystyle =a^2$

$\displaystyle \text{(iii) }\left(a^{-1}+b^{-1}\right)\div\left(a^{-2}-b^{-2}\right)$
$\displaystyle =\left(\frac1a+\frac1b\right)\div\left(\frac1{a^2}-\frac1{b^2}\right)$
$\displaystyle =\frac{a+b}{ab}\div\frac{b^2-a^2}{a^2b^2}$
$\displaystyle =\frac{a+b}{ab}\times\frac{a^2b^2}{(b+a)(b-a)}$
$\displaystyle =\frac{ab}{b-a}$

$\displaystyle \text{(iv) }\left(a^{-1}+b^{-1}\right)\div(ab)^{-1}$
$\displaystyle =\left(\frac1a+\frac1b\right)\div\frac1{ab}$
$\displaystyle =\frac{a+b}{ab}\times ab$
$\displaystyle =a+b$

$\displaystyle \text{(v) }\left(a^{-1}\times b^{-1}\right)\div\left(a^{-1}+b^{-1}\right)$
$\displaystyle =\left(\frac1a\times\frac1b\right)\div\left(\frac1a+\frac1b\right)$
$\displaystyle =\frac1{ab}\div\frac{a+b}{ab}$
$\displaystyle =\frac1{ab}\times\frac{ab}{a+b}$
$\displaystyle =\frac1{a+b}$

$\displaystyle \text{(vi) }(a+b)^{-1}\times\left(a^{-1}+b^{-1}\right)$
$\displaystyle =\frac1{a+b}\times\left(\frac1a+\frac1b\right)$
$\displaystyle =\frac1{a+b}\times\frac{a+b}{ab}$
$\displaystyle =\frac1{ab}$

$\displaystyle \text{(vii) }\frac{a+b+c}{a^{-1}b^{-1}+b^{-1}c^{-1}+c^{-1}a^{-1}}$
$\displaystyle =\frac{a+b+c}{\frac1{ab}+\frac1{bc}+\frac1{ca}}$
$\displaystyle =\frac{a+b+c}{\frac{c+a+b}{abc}}$
$\displaystyle =\frac{a+b+c}{\frac{a+b+c}{abc}}$
$\displaystyle =abc$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Prove that:}$
$\displaystyle \text{(i) }\left(\frac{x^a}{x^b}\right)^{a+b}\left(\frac{x^b}{x^c}\right)^{b+c}\left(\frac{x^c}{x^a}\right)^{c+a}=1$
$\displaystyle \text{(ii) }\left(\frac{x^a}{x^b}\right)^{\frac1{ab}}\left(\frac{x^b}{x^c}\right)^{\frac1{bc}}\left(\frac{x^c}{x^a}\right)^{\frac1{ca}}=1$
$\displaystyle \text{(iii) }\left(\frac{x^a}{x^b}\right)^{a+b-c}\left(\frac{x^b}{x^c}\right)^{b+c-a}$
$\displaystyle \qquad\times\left(\frac{x^c}{x^a}\right)^{c+a-b}=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) LHS}=\left(\frac{x^a}{x^b}\right)^{a+b}\left(\frac{x^b}{x^c}\right)^{b+c}\left(\frac{x^c}{x^a}\right)^{c+a}$
$\displaystyle =x^{(a-b)(a+b)}\times x^{(b-c)(b+c)}\times x^{(c-a)(c+a)}$
$\displaystyle =x^{a^2-b^2+b^2-c^2+c^2-a^2}$
$\displaystyle =x^0$
$\displaystyle =1$
$\displaystyle =\text{RHS}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \text{(ii) LHS}=\left(\frac{x^a}{x^b}\right)^{\frac1{ab}}\left(\frac{x^b}{x^c}\right)^{\frac1{bc}}\left(\frac{x^c}{x^a}\right)^{\frac1{ca}}$
$\displaystyle =x^{\frac{a-b}{ab}}\times x^{\frac{b-c}{bc}}\times x^{\frac{c-a}{ca}}$
$\displaystyle =x^{\frac1b-\frac1a+\frac1c-\frac1b+\frac1a-\frac1c}$
$\displaystyle =x^0$
$\displaystyle =1$
$\displaystyle =\text{RHS}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \text{(iii) LHS}=\left(\frac{x^a}{x^b}\right)^{a+b-c}\left(\frac{x^b}{x^c}\right)^{b+c-a}\left(\frac{x^c}{x^a}\right)^{c+a-b}$
$\displaystyle =x^{(a-b)(a+b-c)}\times x^{(b-c)(b+c-a)}\times x^{(c-a)(c+a-b)}$
$\displaystyle =x^{a^2-b^2-ac+bc+b^2-c^2-ab+ac+c^2-a^2-bc+ab}$
$\displaystyle =x^0$
$\displaystyle =1$
$\displaystyle =\text{RHS}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Prove that:}$
$\displaystyle \frac{a^{-1}}{a^{-1}+b^{-1}}+\frac{a^{-1}}{a^{-1}-b^{-1}}=\frac{2b^2}{b^2-a^2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{a^{-1}}{a^{-1}+b^{-1}}+\frac{a^{-1}}{a^{-1}-b^{-1}}$
$\displaystyle =\frac{\frac1a}{\frac1a+\frac1b}+\frac{\frac1a}{\frac1a-\frac1b}$
$\displaystyle =\frac{\frac1a}{\frac{a+b}{ab}}+\frac{\frac1a}{\frac{b-a}{ab}}$
$\displaystyle =\frac1a\times\frac{ab}{a+b}+\frac1a\times\frac{ab}{b-a}$
$\displaystyle =\frac{b}{a+b}+\frac{b}{b-a}$
$\displaystyle =\frac{b(b-a)+b(a+b)}{(a+b)(b-a)}$
$\displaystyle =\frac{b^2-ab+ab+b^2}{b^2-a^2}$
$\displaystyle =\frac{2b^2}{b^2-a^2}$
$\displaystyle =\text{RHS}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{Prove that:}$
$\displaystyle \frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}$
$\displaystyle \qquad+\frac{1}{1+x^{b-c}+x^{a-c}}=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}$
$\displaystyle \qquad+\frac{1}{1+x^{b-c}+x^{a-c}}$
$\displaystyle =\frac{1}{x^{-a}\left(x^a+x^b+x^c\right)}+\frac{1}{x^{-b}\left(x^b+x^a+x^c\right)}$
$\displaystyle \qquad+\frac{1}{x^{-c}\left(x^c+x^b+x^a\right)}$
$\displaystyle =\frac{x^a}{x^a+x^b+x^c}+\frac{x^b}{x^a+x^b+x^c}+\frac{x^c}{x^a+x^b+x^c}$
$\displaystyle =\frac{x^a+x^b+x^c}{x^a+x^b+x^c}$
$\displaystyle =1$
$\displaystyle =\text{RHS}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{If }abc=1,\text{ prove that:}$
$\displaystyle \frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }abc=1,\quad a=\frac1{bc}\quad\text{and}\quad a^{-1}=bc.$
$\displaystyle \text{LHS}=\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}$
$\displaystyle =\frac{1}{1+\frac1{bc}+\frac1b}+\frac{1}{1+b+\frac1c}+\frac{1}{1+c+bc}$
$\displaystyle =\frac{1}{\frac{bc+1+c}{bc}}+\frac{1}{\frac{c+bc+1}{c}}+\frac{1}{1+c+bc}$
$\displaystyle =\frac{bc}{bc+c+1}+\frac{c}{bc+c+1}+\frac{1}{bc+c+1}$
$\displaystyle =\frac{bc+c+1}{bc+c+1}$
$\displaystyle =1$
$\displaystyle =\text{RHS}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{If }a,b,c\text{ are positive real numbers, show that:}$
$\displaystyle \sqrt{a^{-1}b}\times\sqrt{b^{-1}c}\times\sqrt{c^{-1}a}=1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sqrt{a^{-1}b}\times\sqrt{b^{-1}c}\times\sqrt{c^{-1}a}$
$\displaystyle =\sqrt{\frac ba}\times\sqrt{\frac cb}\times\sqrt{\frac ac}$
$\displaystyle =\sqrt{\frac ba\times\frac cb\times\frac ac}$
$\displaystyle =\sqrt{\frac{abc}{abc}}$
$\displaystyle =\sqrt1$
$\displaystyle =1$
$\displaystyle =\text{RHS}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{If }\frac{9^n\times3^2\times3^n-(27)^n}{3^{3m}\times2^3}=3^{-3},$
$\displaystyle \text{prove that }m-n=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{9^n\times3^2\times3^n-(27)^n}{3^{3m}\times2^3}=3^{-3}$
$\displaystyle \Rightarrow \frac{(3^2)^n\times3^2\times3^n-(3^3)^n}{3^{3m}\times2^3}=3^{-3}$
$\displaystyle \Rightarrow \frac{3^{2n}\times3^2\times3^n-3^{3n}}{8\times3^{3m}}=3^{-3}$
$\displaystyle \Rightarrow \frac{3^{3n+2}-3^{3n}}{8\times3^{3m}}=3^{-3}$
$\displaystyle \Rightarrow \frac{3^{3n}\left(3^2-1\right)}{8\times3^{3m}}=3^{-3}$
$\displaystyle \Rightarrow \frac{3^{3n}(9-1)}{8\times3^{3m}}=3^{-3}$
$\displaystyle \Rightarrow \frac{8\times3^{3n}}{8\times3^{3m}}=3^{-3}$
$\displaystyle \Rightarrow 3^{3n-3m}=3^{-3}$
$\displaystyle \Rightarrow 3(n-m)=-3$
$\displaystyle \Rightarrow n-m=-1$
$\displaystyle \therefore m-n=1$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{If }21168=x^4\times y^3\times z^2,\text{ find the positive integral}$
$\displaystyle \text{values of }x,y\text{ and }z.$
$\displaystyle \text{Answer:}$
$\displaystyle 21168=2^4\times3^3\times7^2$
$\displaystyle \text{Also, }21168=x^4\times y^3\times z^2$
$\displaystyle \therefore x^4\times y^3\times z^2=2^4\times3^3\times7^2$
$\displaystyle \text{Comparing the corresponding powers,}$
$\displaystyle x^4=2^4,\qquad y^3=3^3,\qquad z^2=7^2$
$\displaystyle \therefore x=2,\qquad y=3,\qquad z=7$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{If }1960=2^a\times5^b\times7^c,\text{ find the values of }a,b,c.$
$\displaystyle \text{Hence, calculate the value of }2^{-a}\times5^{-c}\times7^b.$
$\displaystyle \text{Answer:}$
$\displaystyle 1960=2^3\times5^1\times7^2$
$\displaystyle \text{Also, }1960=2^a\times5^b\times7^c$
$\displaystyle \text{Comparing the powers of the distinct primes }2,5\text{ and }7,$
$\displaystyle a=3,\qquad b=1,\qquad c=2$
$\displaystyle \therefore 2^{-a}\times5^{-c}\times7^b$
$\displaystyle =2^{-3}\times5^{-2}\times7^1$
$\displaystyle =\frac{1}{2^3}\times\frac{1}{5^2}\times7$
$\displaystyle =\frac{7}{8\times25}$
$\displaystyle =\frac{7}{200}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Solve: }3^x=\frac13$
$\displaystyle \text{Answer:}$
$\displaystyle 3^x=\frac13$
$\displaystyle \Rightarrow 3^x=3^{-1}$
$\displaystyle \text{Comparing the exponents,}$
$\displaystyle x=-1$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{Solve: }3^{2x+1}=1$
$\displaystyle \text{Answer:}$
$\displaystyle 3^{2x+1}=1$
$\displaystyle \Rightarrow 3^{2x+1}=3^0$
$\displaystyle \text{Comparing the exponents,}$
$\displaystyle 2x+1=0$
$\displaystyle \Rightarrow 2x=-1$
$\displaystyle \therefore x=-\frac12$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{Solve: }\sqrt{\frac ab}=\left(\frac ba\right)^{1-3x}$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{\frac ab}=\left(\frac ba\right)^{1-3x}$
$\displaystyle \Rightarrow \left(\frac ab\right)^{\frac12}=\left(\frac ab\right)^{-(1-3x)}$
$\displaystyle \Rightarrow \left(\frac ab\right)^{\frac12}=\left(\frac ab\right)^{3x-1}$
$\displaystyle \text{Comparing the exponents,}$
$\displaystyle 3x-1=\frac12$
$\displaystyle \Rightarrow 3x=\frac12+1$
$\displaystyle \Rightarrow 3x=\frac32$
$\displaystyle \therefore x=\frac12$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{Solve: }\left(\sqrt[3]{\frac23}\right)^{x-1}=\frac{27}{8}$
$\displaystyle \text{Answer:}$
$\displaystyle \left(\sqrt[3]{\frac23}\right)^{x-1}=\frac{27}{8}$
$\displaystyle \Rightarrow \left(\frac23\right)^{\frac{x-1}{3}}=\left(\frac32\right)^3$
$\displaystyle \Rightarrow \left(\frac23\right)^{\frac{x-1}{3}}=\left(\frac23\right)^{-3}$
$\displaystyle \text{Comparing the exponents,}$
$\displaystyle \frac{x-1}{3}=-3$
$\displaystyle \Rightarrow x-1=-9$
$\displaystyle \therefore x=-8$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{Solve: }\left(\sqrt{\frac35}\right)^{x-1}=\left(\frac{27}{125}\right)^{-1}$
$\displaystyle \text{Answer:}$
$\displaystyle \left(\sqrt{\frac35}\right)^{x-1}=\left(\frac{27}{125}\right)^{-1}$
$\displaystyle \Rightarrow \left(\frac35\right)^{\frac{x-1}{2}}=\left(\frac{3^3}{5^3}\right)^{-1}$
$\displaystyle \Rightarrow \left(\frac35\right)^{\frac{x-1}{2}}=\left(\frac35\right)^{-3}$
$\displaystyle \text{Comparing the exponents,}$
$\displaystyle \frac{x-1}{2}=-3$
$\displaystyle \Rightarrow x-1=-6$
$\displaystyle \therefore x=-5$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{Solve: }9\times3^x=(27)^{2x-5}$
$\displaystyle \text{Answer:}$
$\displaystyle 9\times3^x=(27)^{2x-5}$
$\displaystyle \Rightarrow 3^2\times3^x=\left(3^3\right)^{2x-5}$
$\displaystyle \Rightarrow 3^{x+2}=3^{6x-15}$
$\displaystyle \text{Comparing the exponents,}$
$\displaystyle x+2=6x-15$
$\displaystyle \Rightarrow 5x=17$
$\displaystyle \therefore x=\frac{17}{5}=3\frac25$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{Solve: }2^{5x-1}=4\times2^{3x+1}$
$\displaystyle \text{Answer:}$
$\displaystyle 2^{5x-1}=4\times2^{3x+1}$
$\displaystyle \Rightarrow 2^{5x-1}=2^2\times2^{3x+1}$
$\displaystyle \Rightarrow 2^{5x-1}=2^{3x+3}$
$\displaystyle \text{Comparing the exponents,}$
$\displaystyle 5x-1=3x+3$
$\displaystyle \Rightarrow 2x=4$
$\displaystyle \therefore x=2$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{Solve: }5^{x-3}\times3^{2x-8}=225$
$\displaystyle \text{Answer:}$
$\displaystyle 5^{x-3}\times3^{2x-8}=225$
$\displaystyle \Rightarrow 5^x\times5^{-3}\times3^{2x}\times3^{-8}=225$
$\displaystyle \Rightarrow 5^x\times\frac1{5^3}\times\left(3^2\right)^x\times\frac1{3^8}=225$
$\displaystyle \Rightarrow (5\times9)^x=225\times5^3\times3^8$
$\displaystyle \Rightarrow 45^x=3^2\times5^2\times5^3\times3^8$
$\displaystyle \Rightarrow 45^x=5^5\times3^{10}$
$\displaystyle \Rightarrow 45^x=\left(5\times3^2\right)^5$
$\displaystyle \Rightarrow 45^x=45^5$
$\displaystyle \text{Comparing the exponents,}$
$\displaystyle x=5$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\text{If }2^x=3^y=12^z,\text{ show that }$
$\displaystyle \frac1z=\frac1y+\frac2x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }2^x=3^y=12^z=k.$
$\displaystyle \therefore 2=k^{\frac1x},\qquad3=k^{\frac1y},\qquad12=k^{\frac1z}.$
$\displaystyle \text{Now }12=2^2\times3.$
$\displaystyle \therefore k^{\frac1z}=\left(k^{\frac1x}\right)^2\times k^{\frac1y}$
$\displaystyle \therefore k^{\frac1z}=k^{\frac2x}\times k^{\frac1y}$
$\displaystyle \therefore k^{\frac1z}=k^{\frac2x+\frac1y}$
$\displaystyle \text{Comparing the exponents,}$
$\displaystyle \frac1z=\frac2x+\frac1y$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\text{If }2^x=3^y=6^{-z},\text{ show that }$
$\displaystyle \frac1x+\frac1y+\frac1z=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }2^x=3^y=6^{-z}=k.$
$\displaystyle \therefore 2=k^{\frac1x},\qquad3=k^{\frac1y},\qquad6=k^{-\frac1z}.$
$\displaystyle \text{Now }6=2\times3.$
$\displaystyle \therefore k^{-\frac1z}=k^{\frac1x}\times k^{\frac1y}$
$\displaystyle \therefore k^{-\frac1z}=k^{\frac1x+\frac1y}$
$\displaystyle \text{Comparing the exponents,}$
$\displaystyle \frac1x+\frac1y=-\frac1z$
$\displaystyle \therefore \frac1x+\frac1y+\frac1z=0$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$