Question

The value of $m$ in the expression $\frac{(16)^{2 m+1} \times(64)^5}{(256)^2 \times 4}=(256)^{3 m}$ is___

Answer 1

Given : $\frac{(16)^{2 m+1} \times(64)^5}{(256)^2 \times 4}=(256)^{3 m}$
$ \therefore \frac{\left(2^4\right)^{2 m+1} \times\left(2^6\right)^5}{\left(2^8\right)^2 \times 2^2}=\left(2^8\right)^{3 m} $
$ \therefore \frac{2^{4(2 m+1)} \times 2^{6 \times 5}}{2^{8 \times 2} \times 2^2}=2^{8 \times 3 m}\left[\text { As, }\left(x^m\right)^n=x^{m n}\right] $
$ \therefore \frac{2^{8 m+4} \times 2^{30}}{2^{16} \times 2^2}=2^{24 m} $
$ \therefore \frac{2^{8 m+4+30}}{2^{16+2}}=2^{24 m}\left[\text { As, } x^m x^n=x^{m+n}\right]$
$ \therefore \frac{2^{8 m+34}}{2^{18}}=2^{24 m}$
$ \therefore 2^{(8 m+34)-18}=2^{24 m}\left[\text { As, } \frac{x^m}{x^n}=x^{m-n}\right] $
$ \therefore 2^{8 m+16}=2^{24 m} $
$ \therefore 8 m+16=24 m \left[\text { As, } a^x=a^y\right. $ $ \text { then } x=y \text { ] } $
$ \therefore 16 m=16 $
$ \therefore m=1 $