How to solve these logarithmic equations? I want to solve: 


*

*$n^k = n^{log n}$ where k is real number for $n$.

*$a^n = n^{log n}$ where a is real number for $n$.
But I fail to do both of them. Can you tell me how to solve it?
 A: For the first equation, using what   André Nicolas suggested makes the problem quite simple.
The second one is more difficult and I do not think that analytical solutions could be obtained. Considering $$a^n = n^{\log n}$$ cen be transformed to $$n \log(a)=\log^2(n)$$ So, let us consider the function $$f(n)=\frac{\log^2(n)}n$$ and compute its derivative $$f'(n)=\frac{2 \log (n)}{n^2}-\frac{\log ^2(n)}{n^2}=-\frac{(\log (n)-2) \log (n)}{n^2}$$ It cancels for $n=1$ and $n=e^2$. So, the function starts at $+\infty$, decreases to $f(1)=0$, then increases up to $f(e^2)=\frac{4}{e^2}$ and then decreases to $0$.
So, 


*

*if $\log(a) \gt\frac{4}{e^2}$, there is one root to the equation (smowhere between $0$ and $1$). 

*if $0 <\log(a) \lt\frac{4}{e^2}$, there are three roots (one of them between $0$ and $1$, another one between $1$ and $e^2$, another one larger than $e^2$)

*if $\log(a)=0$, there is only one root corresponding to $n=1$

*if $\log(a) =\frac{4}{e^2}$, there is one root between $0$ and $1$ plus one root corresponding to $n=e^2$

*if $\log(a)<0$, no root


To find the roots, numerical methods such as Newton would be required.
A plot of the function would clearly show all the above.
A: For a solution to the first part by inspection or equating exponents,
$$ n = e^k,1 $$
As pointed by Rory Daulton $ n=1 $ is included as a solution because $ 1^{any\, real\,number}= 1. $
For the second part $n$ cannot be expressed in terms of $a$ analytically ( closed form) due to  transcendental dependence, but  $a$ can  be expressed in terms of $n$ whose inverse function is required.
$$    a = {(n^{log n} )}^{1/n} $$
It can be numerically solved for a given $a$ in real regime.
