How many solutions: $ a^{x} = x $ I'm clueless.
For $ a > 0 $ how many solutions: $ a^{x} = x $ 
 A: Since $a^x>0$, for all $x$, if a solution exists it is positive. So it's not restrictive to assume $x>0$. Then the equation becomes the same as
$$
x\log a=\log x
$$
(natural logarithm). Let's look at $a>1$, to begin with.
Consider the function $f(x)=x\log a-\log x$. It's easy to see that
$$
\lim_{x\to0}f(x)=\infty,
\qquad
\lim_{x\to\infty}f(x)=\infty
$$
Thus the function has an absolute minimum. Since
$$
f'(x)=\log a-\frac{1}{x}
$$
the minimum is at $\frac{1}{\log a}$. Now
$$
f\left(\frac{1}{\log a}\right)=1+\log\log a
$$
and we have


*

*no solution if $1+\log\log a>0$, that is, $a>e^{1/e}\approx 1.44467$;

*one solution if $1+\log\log a=0$, that is, $a=e^{1/e}$;

*two solutions if $1+\log\log a<0$, that is, $1<a<e^{1/e}$.


The case $a=1$ is obvious.
In the case $0<a<1$, the function $f$ has
$$
\lim_{x\to0}f(x)=\infty,
\qquad
\lim_{x\to\infty}f(x)=-\infty
$$
and $f'(x)<0$ for all $x>0$, so the equation has only one solution.

Alternative method. The equation, with the same trick, is equivalent to
$$
\frac{\log x}{x}=\log a
$$
Let's consider the function $g(x)=\frac{\log x}{x}$. Then
$$
\lim_{x\to 0}g(x)=-\infty,
\qquad
\lim_{x\to\infty}g(x)=0
$$
Since
$$
g'(x)=\frac{1-\log x}{x^2}
$$
we see that $g$ has a maximum at $x=e$; moreover $g(e)=\frac{1}{e}$.

Since $\log a$ takes on every real value exactly once, we see we have


*

*no solution if $\log a>1/e$

*one solution if $\log a=1/e$

*two solutions if $0<\log a<1/e$

*one solution if $\log a\le 0$

A: Assume $a>0$
$$a^x=x\Longleftrightarrow\ln\left(a^x\right)=\ln(x)\Longleftrightarrow$$
$$x\ln\left(a\right)=\ln(x)\Longleftrightarrow\ln\left(a\right)=\frac{\ln(x)}{x}\Longleftrightarrow$$
$$a=e^{\frac{\ln(x)}{x}}\Longleftrightarrow a=\exp\left[\frac{\ln(x)}{x}\right]\Longleftrightarrow a=\sqrt[x]{x}$$
