How many solutions has $x^\alpha =\ln x$?

Question

I have to determine the number of real solutions of $x^\alpha = \ln x$ on varying of $\alpha > 0$.

How would you proceed?

You can have a solution in terms of Lambert W function. – Mhenni Benghorbal Aug 29 '13 at 10:37 — Mhenni Benghorbal, Aug 29 '13 at 10:37

score 3 · Answer 1 · answered Aug 29 '13 at 10:51

3

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.

answered Aug 29 '13 at 10:51

Arash

11,131

1

+1 Thanks a lot! I have accepted another answer ONLY because it is as complete as yours and it was given some seconds before. – collimarco Aug 29 '13 at 11:44

score 1 · Answer 2 · edited Apr 13 '17 at 12:20

1

Here is a solution

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

See here for related problem.

edited Apr 13 '17 at 12:20

Community

1

answered Aug 29 '13 at 10:40

Mhenni Benghorbal

47,431

score 1 · Accepted Answer · answered Aug 29 '13 at 10:51

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.

How many solutions has $x^\alpha =\ln x$?

3 Answers3