Question

a^mn=a^m^n then find m

Answer 1

$\textbf{Given:}$

$\bf\,a^{mn}=a^{m^n}$

$\text{Equating powers on both sides,we get}$

$mn=m^n$

$\text{Taking logarithms on both sides,we get}$

$log\;mn=log\;m^n$

$\implies\;log\;m+log\;n=n\;log\;m$

$\implies\;log\;m+log\;n=log\;m^n$

$\implies\;log\;m-log\;m^n=log\;n$

$\implies\;log\frac{m}{m^n}=log\;n$

$\implies\;m^{1-n}=n$

$\implies\,m=n^(\frac{1}{1-n})$

Answer 2

The value of $m$ is $n^{\frac{1}{1-n} }$

Step-by-step explanation:

Given,

$a^{mn} =a^{m^{n} }$

Taking logarithms on both sides,

⇒$log(a^{mn} )=log(a^{m^{n} })$

⇒$mn.log(a)=m^{n}.log(a)$  (∵$loga^{x} =x.loga$ )

⇒ $mn=m^{n}$ (By canceling $log(a)$ on both sides )

⇒ $log(m)+log(n)=log(m^{n})$ (Taking logarithms on both sides)

⇒ $log(m)-log(m^{n})=log(n)$

⇒$log(\frac{m}{m^{n} } )=log(n)$

⇒$m^{1-n} =n$

∴ $m=n^{\frac{1}{1-n} }$

So, The value of $m$ is $n^{\frac{1}{1-n} }$