What values of $a$ make $x=a^x$ solvable? I was wondering what the solution to $x=e^x$ was, but then I graphed $y=x$ and $y=e^x$ and saw that they didn't intersect. I assume they wouldn't intersect for a base greater than $e$ either. So, I wanted to know what values of $a$ would give the equation $x=a^x$ a solution. The first thing you notice is that $x>0$ because $x=a^x>0$. Also, the values $a\in(0,1)$ should work based on the shape of exponential decay (and $a=1$ of course), but the challenge is for $a>1$. The conclusion I reached was that the rest of the values for $a$ should be inside $(1,e)$. My reasoning uses the fact that $x>\ln(x)$ on the domain of $\ln(x)$.
If you rearrange $x=a^x$ you get $$x=a^x$$ $$\ln(x)=\ln(a^x)$$ $$\ln(x)=x\ln(a)$$ $$\frac{\ln(x)}{\ln(a)}=x$$ In order for this equation to be unsolvable, the function on the left must never intersect the function on the right, so one must always be greater. Because $\ln(x)$ grows more slowly, x would have to be the greater function. This leads to the inequality $$x>\frac{\ln(x)}{\ln(a)}$$ $$x\ln(a)>\ln(x)$$ This inequality is automatically true if $x\ln(a)>x$. Dividing on both sides gives $ln(a)>1\rightarrow a>e$. Therefore when $a>e$ the equation can't be solved. But what subinterval of $(1,e)$ contains the a-values I'm looking for?
 A: If you consider the graph of $y=x$ and you start decreasing $a$ in the graph of $y=a^x$ (say, starting from $a=e$) you can see that there will be a value for $a$ when the exponential curve just touches, and is tangent to, $y=x$. This is the maximal value of $a$ where there will be a solution. For $a$ less than that (yet greater than $1$) there will be two solutions because that exponential curve is concave up and both functions are increasing. You have already worked out that there is a solution for $a$ in $(0,1]$.
To find that maximal value of $a$, you have two conditions (from the intersection and the tangency):
$$x=a^x\qquad 1=a^x\ln(a)$$
So $1=x\ln(a)$, and therefore $x=\frac{1}{\ln(a)}$. From this, $$\begin{align}
1&=a^{1/\ln(a)}\ln(a)\\
1&=\left(e^{\ln(a)}\right)^{1/\ln(a)}\ln(a)\\
1&=e\ln(a)
\end{align}$$
From which $a=e^{1/e}$. So there are solutions for $a\in(0,e^{1/e}]$.
Specifically, there is precisely one solution when $a\in(0,1]\cup\{e^{1/e}\}$, and precisely two solutions when $a\in(1,e^{1/e})$.
A: First, let's solve this equation. We know that
\begin{align*}
x&=a^x\\
&=\exp\left(x\ln(a)\right)\\
\implies x\exp\left(-x\ln(a)\right)&=1\\
\implies-\ln(a)x\exp\left(-x\ln(a)\right)&=-\ln(a)\\
-\ln(a)x&=W(-\ln(a))\\
x&=-\frac{W(-\ln(a))}{\ln(a)}
\end{align*}
The graph of the solution looks like this

We now want to find the interval from 0 to somewhere before 1.5 where real solutions exist.
The restrictions arising from the $\ln(x)$ function merely state for a nonnegative interval. This gives us our lower bound.
For the upper bound, we must note that the product log is imaginary for all values to the left of $x=-\frac 1e$. Thus, to find the upper restriction we just set
$$-\ln(a)=-\frac 1e$$
to which the solution is $a=\sqrt[e]{e}$.
Hence, the answer to your question is the interval $$a\in(0, \sqrt[e]{e}]$$
A: Consider $f(x)=a^x-x$ for $a>1$
$f'(x)= (\ln a)a^x -1$
$f'(x)= 0$ when $x=-\frac{\ln (\ln a)}{\ln a}$
We can show this is a minimum point by consider the sign of $f'(x)$
At minimum point, $f(x)=a^x-x=\frac{1}{\ln a}+\frac{\ln (\ln a)}{\ln a}=\frac{1+\ln(\ln a)}{\ln a}$
For $a>e^{\frac{1}{e}}$, $f(x)>0$ for all x.
For $1<a<e^{\frac{1}{e}}$, $f(c)<=0$ at minimum point, so there exist a root. Since $f(0)=1$
