How many solutions has $x^\alpha =\ln x$? I have to determine the number of real solutions of $x^\alpha = \ln  x$ on varying of $\alpha > 0$.
How would you proceed?
 A: For $\alpha>0$, you can see that $g(x)=x^\alpha-lnx$ is a continuous function in $(0,\infty)$. By taking a derivative you can see that the function has one minimum which is $\frac{1}{\alpha}(1+\ln(\alpha))$. Now if minimum is negative, it has two real root, if it is positive there is no root and for zero-minimum there is one root and it is the minimum point. $\alpha=\frac{1}{e}$ is the point below which there is two roots. 
A: Here is a solution 

$$\large {{\rm e}^{-{\frac {{W_k} ( -\alpha ) }{\alpha}}}}. $$

See here for related problem.
A: Some hints:
$$f(x):=x^\alpha-\log x\implies f'(x)=\alpha x^{\alpha-1}-\frac1x\ge0\iff\alpha x^\alpha\ge1\iff$$
$$\iff x\ge\frac1{\alpha^{1/\alpha}}\;\;\;\text{(we have}\;\;x,\alpha>0)$$
We thus see $\;f(x)\;$ is monotone descending for $\;x< \alpha^{-1/\alpha}\;$ and zero or ascending otherwise, thus
$$f\left(\alpha^{-1/\alpha}\right)=\frac1\alpha+\log\alpha$$
Observe though that
$$\lim_{x\to 0^+}f(x)=+\infty\;,\;\;\lim_{x\to\infty} f(x)=+\infty\;,\;\;\text{so:}$$
If the above value is positive there are zero solutions, if it is zero...etc.
