If m^n = n^m,then the value of m in terms of | vedclass

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(C) Given the equation $m^n = n^m$. Taking the natural logarithm on both sides: $\ln(m^n) = \ln(n^m)$ $n \ln(m) = m \ln(n)$ Divide both sides by $mn$:

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

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(C) Given the equation $m^n = n^m$. Taking the natural logarithm on both sides: $\ln(m^n) = \ln(n^m)$ $n \ln(m) = m \ln(n)$ Divide both sides by $mn$: $\frac{\ln(m)}{m} = \frac{\ln(n)}{n}$ Let $f(x) = \frac{\ln(x)}{x}$. For $m 
eq n$,the equation $f(m) = f(n)$ holds for specific values. If we set $m = n^k$,then $(n^k)^n = n^{(n^k)}$. $n^{kn} = n^{n^k}$. Equating the exponents: $kn = n^k$. $k = n^{k-1}$. Taking the $k$-th root: $k^{1/k} = n^{1/n}$ is not directly helpful,but solving $m = n^{1/(n-1)}$ is a standard result for this transcendental equation.

If $m^n = n^m$,then the value of $m$ in terms of $n$ is:

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