Solutions of $a^x = x$

Question

How can I find a bound for the solutions of the following equation without using the Lambert function?

$$a^x = x,$$ where $a \in \mathbb{R}$.

Answer 1

I am assuming you are only looking for real solutions. Even if not, this answer can still be useful. We can see that for there to be a real solution, $a$ must be in the form $x^\frac{1}{x}$. Using calculus to maximize this gives that the maximum value of $x^\frac{1}{x}$ and thus the maximum value of $a$ is $e^\frac{1}{e}$. From this you get that the real solutions for $x$ have to be less than or equal to $e$.

For a lower bound, we can again use the same equation, noting that minimum value of $x^\frac{1}{x}$ approaches 0, and hence the lower bound for $x$ is 0.

Answer 2

There are no algebraic "closed-form" expressions for the solutions. However, we can rewrite it as $$\begin{align*} -a^x&=-x\\ -1&=(-x)a^{(-x)}\\ -1&=(-x)e^{\ln(a)\cdot (-x)}\\ -\ln(a)&=(-\ln(a)x)e^{(-\ln(a)x)}\\ -\ln(a)&=ye^y,\quad\text{where } y=-\ln(a)x. \end{align*}$$ The solution of this equation is, by the definition of the Lambert W function, $$y=W(-\ln(a)),$$ which is to say $$x=-\frac{W(-\ln(a))}{\ln(a)}.$$