Find the value of $a$ where $a^x = \frac{\log(x)}{\log(a)}$ has only one solution I have those two functions
$f(x) = a^x$
$g(x) = \frac{\log(x)}{\log(a)}$
Where $g(x)$ is symmetric to $f(x)$ w.r.t. the $y=x$ axis.
With $a = 1.3$ we have two solutions to $f(x) = g(x)$:

With a = 1.5 we have no solutions to $f(x) = g(x)$:

So there is a value $a$ (between 1.3 and 1.5) where $f(x) = g(x)$ admit one and only one solution. Graphically I've noticed that those value should be close to 1.4446 but I cannot find a more precise answer. I don't think that it is possible to isolate $x$ in the equation $f(x) = g(x)$ so how could I determine this value ?
 A: Since the graph of $g$ is the mirror image of the graph of $f$ through the line $y=x$, we have $f(x)=g(x)$ if and only if $f(x)=x$. So we want $f(x)=x$ to have only one solution. Since we have $f(x)-x\to\infty$ when $x\to\pm\infty$, we need to have $f'(x)=1$ at the point where $f(x)=x$.
Note that $f'(x)=\log(a)\cdot a^x$, so solving $f'(x)=1$ yields $x=\log_a(1/\log(a))$. Then the equation $f(x)=x$ becomes $$\frac1{\log(a)}=\log_a\left(\frac1{\log(a)}\right)=\frac{-\log(\log(a))}{\log(a)},$$ so $-\log(\log(a))=1$, so $a=e^{e^{-1}}\approx1.44466786$.
A: Consider the function
$$a^x-\frac{\log x}{\log a}=e^{x\ln a}-\frac{\log x}{\log a}\tag1$$
It reaches the minimum where its derivative is zero:
$$x=\frac{W(1/\log a)}{\log a}$$
(Here $W$ is the Lambert W-function.) We now substitute this back into $(1)$ and get what appears to be a monstrosity. We want this to be zero as well.
$$e^{W(1/\log a)}-\frac{\log\frac{W(1/\log a)}{\log a}}{\log a}=0$$
But now, if we substitute $a=e^{1/e}$, $W(1/\log a)=W(e)=1$ and we get
$$e^{1}-\frac{\log\frac{1}{\log e^{1/e}}}{\log e^{1/e}}=e-\frac{\log\frac{1}{1/e}}{1/e}=e-\frac{1}{1/e}=0$$
So in fact, the correct $a$ is $e^{1/e}$.
