Basic of Logarithms
easy

If ${({a^m})^n} = {a^{{m^n}}}$, then the value of $'m'$ in terms of $'n'$ is

A

$n$

B

${n^{1/m}}$

C

${n^{1/(n - 1)}}$

D

None of these

Solution

(c) ${({a^m})^n} = {a^{{m^n}}} = {a^{mn}} \Rightarrow mn = {m^n}$

$ \Rightarrow $$n = {m^{n – 1}} \Rightarrow m = {n^{{1 \over {n – 1}}}}$.

Standard 11
Mathematics

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