Find solutions of $\alpha x^n = \ln x$ How can I find the solutions of $$\alpha x^n = \ln x$$ when $\alpha \in \mathbb{R}$ and $n\in \mathbb{Q}$? Or, if it is not possible to have closed form solutions, how can I prove that there exist one (or there is no solution) and that it is unique? (I'm particularly interested in the cases $n=2$, $n=1$, and $n=1/2$).
 A: There's no point in restricting $n$ to be rational.
Set $x^n=t$, so $x=t^{1/n}$ and the equation becomes
$$
\alpha t=\frac{1}{n}\log t
$$
or $n\alpha t=\log t$ and we can so study the problem
$$
kt=\log t
$$
Consider the function
$$
f(t)=kt-\log t
$$
defined for $t>0$. We have
$$
\lim_{t\to0}f(t)=\infty
$$
and
$$
\lim_{t\to\infty}f(t)=
\begin{cases}
\infty & \text{if $k>0$}\\
-\infty & \text{if $k\le 0$}
\end{cases}
$$
(check it).
The derivative is
$$
f'(t)=k-\frac{1}{t}=\frac{kt-1}{t}
$$
which is everywhere negative if $k\le0$, so in this case the equation has one solution.
For $k>0$, the minimum is attained at $1/k$ and $f(1/k)=1+\log k$. So we have


*

*no solution if $k>e^{-1}$,

*one solution if $k=e^{-1}$,

*two solutions if $0<k<e^{-1}$.


Since $k=n\alpha$, it's easy to translate the result in terms of $\alpha$ in the cases $n=2$, $n=1$ and $n=1/2$.
A: In the case that $\alpha n\not =0$, we have
$$\alpha x^n = \ln x$$
$$ \alpha nx^{n}= n\ln x$$
$$ \alpha nx^{n}= \ln x^{n}$$
$$ e^{\alpha nx^{n}} = x^{n}$$
$$ 1= \frac{x^{n}}{e^{\alpha nx^{n}}}$$
$$ 1= x^{n}e^{-\alpha nx^{n}} $$
$$ -\alpha n= -\alpha nx^{n}e^{-\alpha nx^{n}} $$
$$ W(-\alpha n)= -\alpha nx^{n} $$
$$ -\frac{W(-\alpha n)}{\alpha n}= x^{n} $$
$$ x=\left(-\frac{W(-\alpha n)}{\alpha n}\right)^{\frac1n}$$
Where $W$ is the Lambert W function.
