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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