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

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Solution

$m \times n = m^{n}$
Taking logarithm to the base $m$ on both sides
${log}_{m} m n = {log}_{m} m^{n}$
${log}_{m} m + {log}_{m} n = n {log}_{m} m$
$1 + {log}_{m} n = n$
${log}_{m} n = n - 1$
$m^{n - 1} = n$
$∴ m = n^{\frac{1}{n - 1}}$

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

$m \times n = m^{n}$
Taking logarithm to the base $m$ on both sides
${log}_{m} m n = {log}_{m} m^{n}$
${log}_{m} m + {log}_{m} n = n {log}_{m} m$
$1 + {log}_{m} n = n$
${log}_{m} n = n - 1$
$m^{n - 1} = n$
$∴ m = n^{\frac{1}{n - 1}}$