Question

Evaluate:- $(\frac{27}{125}{)}^{\frac{2}{3}}\times (\frac{9}{25}{)}^{\frac{-3}{2}}$

• A. $\frac{5}{3}$
• B. $\frac{3}{5}$
• C. $\frac{5}{4}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A

$\frac{5}{3}$

$(\frac{27}{125}{)}^{\frac{2}{3}}\times (\frac{9}{25}{)}^{\frac{-3}{2}}$

$=(\frac{3^3}{5^3}{)}^{\frac{2}{3}}\times (\frac{3^2}{5^2}{)}^{\frac{-3}{2}}$

$=\left(\right(\frac{3}{5}{)}^3{)}^{\frac{2}{3}}\times \left(\right(\frac{3}{5}{)}^2{)}^{\frac{-3}{2}}$
$[\because (\frac{a}{b}{)}^m=\frac{a^m}{b^m}]$

$=(\frac{3}{5}{)}^{3\times \frac{2}{3}}\times (\frac{3}{5}{)}^{2\times \frac{-3}{2}}$
$[\because ((a{)}^m{)}^n=a^{(m\times n)}]$

$=(\frac{3}{5}{)}^2\times (\frac{3}{5}{)}^{-3}$

$=(\frac{3}{5}{)}^{-1}$
$[\because a^m\times a^n=a^{(m+n)}]$

$=\frac{5}{3}$
$[\because a^{-m}=\frac{1}{a^m}]$