Question

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

Answer 1

Answer: 1

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