For what value of $x$ is $x^b = a^x$

Question

For what values of $x$ is $x^b = a^x$, where $ b = a^2$. I was looking at a specific case where $a = 2$, but I don't know how to find the result. It tried using logs but it doesn't seem to help.

Answer 1

Using logarithms/elementary math functions will not solve this. The solution will come from using the Lambert-W function.

Using logarithms will solve problems for one side, but create problems for the other. This will keep on occurring, so logarithms cannot fully help you.

You can only get up till this point:$$x=\log_a (x^b)$$

For simplicity, just assume that $a=e$ (Euler's number).

By definition, $v=ue^u \implies W(v) = u$. Let $W$ be the Lambert-W function. $$\ln(x^b) = x$$ $$x^b = e^x$$ $$x = e^{x/b}$$ $$1 = xe^{-x/b}$$ $$\dfrac{-1}{b} = \dfrac{-x}{b}e^{-x/b}$$ $$W\left(\dfrac{-1}{b}\right) = \dfrac{-x}{b}$$ $$-bW\left(\dfrac{-1}{b}\right)=x$$