If $\left(\frac{3}{4}\right)^{-6} \times \left(\frac{4}{3}\right)^{-8} = \left(\frac{3}{4}\right)^n$, then the value of n is- Join @Apnapdfs

Explanation

## Page 43 $$ \left(\frac{3}{4}\right)^{-6} \times \left(\frac{4}{3}\right)^{-8} = \left(\frac{3}{4}\right)^n $$ $$ \left(\frac{3}{4}\right)^{-6} \times \left(\frac{3}{4}\right)^{8} = \left(\frac{3}{4}\right)^n $$ $$ \left(\frac{3}{4}\right)^{8-6} = \left(\frac{3}{4}\right)^n $$ n = 2 $$ = \frac{2\{a^2 + b^2 + 2ab - 2ab\}}{(a^2 + b^2)} $$ $$ = \frac{2(a^2 + b^2)}{(a^2 + b^2)} $$ $$ = 2 $$ ## Surds, Indices and Square Root, Cube Root