Solve $x$ in $x = A^{x}$ for given $A$.

Question

I can see from graphing the problem that the equation $x = A^x$ will only have solutions when $A < 1.446...$ (approx).

For some value of $A$, say $k$, it will have exactly one solution, and then for $A < k$ it will have $2$ solutions.

How can we work out $k$? And how can we solve for the two values of $x$ given a specific $A < k$?

Look at Lambert function – Claude Leibovici Jun 23 '22 at 08:08 — Claude Leibovici, Jun 23 '22 at 08:08
Compare with this post. – Dietrich Burde Jun 23 '22 at 08:10 — Dietrich Burde, Jun 23 '22 at 08:10

Claude Leibovici · Accepted Answer · 2022-06-24T02:38:22.073

Without any special function, consider that you look for ther zero of $$f(x)=A^x-x$$ It is simpler to take logarithms (assuming $x>0$) and consider $$g(x)=x \log(A)-\log(x)$$ $$g'(x)=\log(A)-\frac 1 x \qquad \text{and} \qquad g''(x)=\frac 1 {x ^2} ~>0~\forall x$$

The minimum of the function occurs at $$x_*=\frac 1{\log(A)}\implies g(x_*)=\log (\log (A))+1$$ In order to have two roots you need that $$\log (\log (A))+1 <1 \implies A < e^{\frac{1}{e}}=1.44467$$

Now, if $A< e^{\frac{1}{e}}$, the only explicit solutions are given in terms of Lambert function which is not elementary but simple. This function has a lot of applications (to give you an idea, just type Lambert of the search bar : $3932$ entries).

In the real domain the small and large solutions are $$x_1=-\frac{W_0(-\log (A))}{\log (A)} \qquad \text{and}\qquad x_2=-\frac{W_{-1}(-\log (A))}{\log (A)} $$ In the linked page, you will find a series of approximations for their computation.

Thank you very much Claude! This was incredibly helpful. – mrsquee Jun 23 '22 at 08:32 — mrsquee, Jun 23 '22 at 08:32

Solve $x$ in $x = A^{x}$ for given $A$.

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