Question

If $\left(\frac{a}{b}\right)={\left(\frac{5}{2}\right)}^{-3}\times {\left(\frac{8}{15}\right)}^{-3}$, then ${\left(\frac{a}{b}\right)}^{-2}$ is equal to

• A. ${\left(\frac{4}{3}\right)}^6$
• B. ${\left(\frac{3}{4}\right)}^{-6}$
• C. Both (A) and (B)
• D. None of the above

Answer 1

Given:

$\left(\frac{a}{b}\right)={\left(\frac{5}{2}\right)}^{-3}\times {\left(\frac{8}{15}\right)}^{-3}$

$\Rightarrow \left(\frac{a}{b}\right)={\left(\frac{2}{5}\right)}^3\times {\left(\frac{15}{8}\right)}^3$

$\Rightarrow \frac{a}{b}=\frac{2}{5}\times \frac{2}{5}\times \frac{2}{5}\times \frac{15}{8}\times \frac{15}{8}\times \frac{15}{8}$

$\Rightarrow \frac{a}{b}=\frac{1}{1}\times \frac{1}{1}\times \frac{1}{1}\times \frac{3}{4}\times \frac{3}{4}\times \frac{3}{4}$

$\Rightarrow \frac{a}{b}={\left(\frac{3}{4}\right)}^3$

$\Rightarrow {\left(\frac{a}{b}\right)}^2={\left({\left(\frac{3}{4}\right)}^3\right)}^2$

$\Rightarrow {\left(\frac{a}{b}\right)}^2={\left(\frac{3}{4}\right)}^6$ ---(1) [Since, $(a^m{)}^n=a^{m\times n}$]

$\Rightarrow {\left(\frac{a}{b}\right)}^2={\left(\frac{4}{3}\right)}^{-6}$ ---(2) [Since, ${\left(\frac{a}{b}\right)}^m={\left(\frac{b}{a}\right)}^{-m},a,b\ne 0$]

$\therefore {\left(\frac{a}{b}\right)}^{-2}={\left(\frac{4}{3}\right)}^{-2\times -3}={\left(\frac{4}{3}\right)}^6={\left(\frac{3}{4}\right)}^{-6}$

The correct option is C. Both (A) and (B)