Proving $x^{\log n} = n^{\log x}$

Question

How could one go about proving that $$x^{\log n} = n^{\log x}\ {?}$$ I'm not really sure how to get started.

Answer 1

Notice, $$x^{\log(n)}=n^{\log (x)}$$ Taking log on base $x$, we get

$$\log_x(x^{\log(n)})=\log_x(n^{\log (x)})$$ $$\log(n)\log_x(x)=\log (x)\log_x(n)$$ $$\log(n)=\log (x)\log_x(n)$$ $$\frac{\log(n)}{\log (x)}=\log_x(n)$$ $$\log_x(n)=\log_x(n)$$ Hence, proved

Answer 2

Hint: $$ x^{\log(N)} = \left( 10^{\log(x)} \right)^{\log(N)} = 10^{\log(x)\log(N)} $$ Alternatively, try taking the log of both sides.

Answer 3

You can begin with $$(\log N)(\log x)=(\log x)(\log N)$$ Then \begin{align} e^{(\log N)(\log x)}&=e^{(\log x)(\log N)}\\ (e^{\log N})^{\log x}&=(e^{\log x})^{\log N}\\ N^{\log x}&=x^{\log N} \end{align}

Answer 4

Let $\log_an=N,\log_ax=X\implies n=a^N,x=a^X$

$x^{\log n}=(a^X)^N=a^{XN}$

$n^{\log_ax}=?$