How to solve x^x=2x

Question

I dont know how to solve this equation.

$$x^x=2x$$

I tried to use Lambert W fuction. But it didnt work.

Answer 1

Whenever you see an $x$ that you want to bring down, take the $\log$: first, observe that $x>0$ for $x^x$ to be well-defined.

In fact, $\log(x^x) = x \log(x)$ and hence

$$x \log(x) = \log(2x)$$ $$x \log(x) = \log(2)+\log(x)$$ $$(x-1)\log(x) = \log(2)$$

And now you can look at the graph of the function $f(x)=(x-1)\log(x)$. Clearly $$\lim_{x \to +\infty} f(x) = +\infty\quad , \quad \lim_{x \to 0^+} f(x) = +\infty$$

Since $f'(x) = -\frac{1}{x} + \log(x) +1 $ vanishes in a single point ($x=1$), there is a unique local minimum when $x=1$. Compute $f(1)=0$ and observe that $f(1)<\log(2)$.

This means that the equation $f(x)=\log(2)$ has two solutions. One happens to be $x=2$, the other can be determined via numerical methods.

Answer 2

As @AnalysisStudent0414 showed in his/her answer, beside the trivial $x=2$, there is another solution in $(0,1)$.

Build the series expansion $$x^x-2x=1+x (\log (x)-2)+O\left(x^2\right)$$ Ignoring the higher order terms, the approximate solution is $$x_0=-\frac{1}{W_{-1}\left(-\frac{1}{e^2}\right)}\approx 0.317844$$ while the exact solution, obtained using Newton method, is $0.346323$.

Using Newton, Halley or Householder method, the solution is obtained in a couple of iterations.