The equations $a^x=\log_ax$ 
How many solutions does equation
  $$a^x=\log_ax$$

1) Let $a>1$
For example: $a=2$
There are no solutions 

2) Let $0<a<1$
For example: $a=\frac12$
There are one solution here.

But if $a=1,1$ then

 A: The equation looks absolutely awful in this form, but the symmetry and the images actually help you find a way of transforming the equation differently. You know the solutions will always lie on the diagonal $y=x$, so instead of solving your equation, you can just solve
$$a^x=x$$
This is a transcendental equation, but is in a very well known form. Equations which have exponentials and polynomials at the same time, can sometimes be reduced to the Lambert W function: a transcendental function that is defined as the solution of the equation $x e^x=y$ as such: $x=W(y)$.
To get this form, turn it around:
$$1=x a^{-x}= x e^{-x \ln a}$$
Now use a new variable $u=-x \ln a$, and you get
$$-\ln a = u e^u$$
which has a solution
$$ u= W(-\ln a)$$
How is this useful? We know, that this will have a solution, when $W$ has a solution. $W(y)$ has two solutions, when $-1/e<y<0$, and one solution for $y>0$. It has no solutions for $y<-1/e$.
From this, you can get the ranges for $a$.
The midpoint between $1$ solution and $2$ solutions is $-\ln a =0$, which corresponds to $a=1$ (this is the midpoint between rising exponentials, which intersect the diagonal twice, and falling exponentials which have just 1 intersection). The critical point where it gets to no solutions at all, is at
$$-\ln a = -1/e\rightarrow a=e^{1/e}\approx 1.44466786$$
The final answer is therefore:


*

*1 solution if $0<a\leq 1$

*2 solutions if $1<a<e^{1/e}$

*1 double solution (touching curves) if $a=e^{1/e}$

*no solutions for $a>e^{1/e}$

A: You can instead solve $x=\log_ax$, because $a^x$ and $\log_ax$ are inverse function to one another.
Since $\log_ax=\frac{\log x}{\log a}$ (natural logarithm), we can as well consider $x\log a=\log x$. Consider
$$
f(x)=x\log a-\log x
$$
over $(0,\infty)$; then $\lim_{x\to0}f(x)=\infty$ and
$$
\lim_{x\to\infty}f(x)=
\begin{cases}
-\infty & 0<a<1 \\[4px]
\infty & a>1
\end{cases}
$$
Moreover
$$
f'(x)=\log a-\frac{1}{x}
$$
For $0<a<1$ the derivative is everywhere negative and the equation $x=\log_ax$ has a single solution.
For $a>1$, the function is decreasing over $(0,1/\log a]$ and increasing over $[1/\log a,\infty)$. The minimum value is
$$
f(1/\log a)=1+\log\log a
$$
which is negative if and only if $\log\log a<-1$, that is, $\log a<e^{-1}$ and so $a<e^{1/e}\approx1.444667861$.
The equation has


*

*one solution for $0<a<1$ (the function $f$ is decreasing)

*two solutions for $1<a<e^{1/e}$ (the function $f$ has negative minimum)

*one solution for $a=e^{1/e}$ (the function $f$ has minimum $0$)

*no solution for $a>e^{1/e}$ (the function $f$ has positive minimum)

A: some corrections:
for example:
a=0.027

three solutions.





a
number
tangent line




$a>\sqrt[e]{e}$
0



$a=\sqrt[e]{e}$
1
$y=x$


$1 < a <\sqrt[e]{e}$
2



$ \dfrac 1{e^e}< a < 1$
1



$ a=\dfrac 1{e^e} $
1
$y=\dfrac 2e-x$


$0< a <\dfrac 1{e^e}$
3





link: https://www.geogebra.org/calculator/rqg75ze2
