$x$ , $a$ ∈ $ℝ$
$x$ > $0$ , $a \neq 1$
Is there any $x$ that makes the below equation have only one repeated root? How about other states? [ From root we mean root of x with a given a ]
For example a $x$ that makes the equation have two simple roots or maybe no roots?
Equation :
$$\log_ax = a ^ x$$
(It is good to note that the x can be generalized to any other function.)
We need to assume $a>0$, otherwise the functions are not defined.
Let $f_a(x) = a^x$. Let's note that $\log_a x = f_a^{-1}(x)$. We have therefore equation $$ f_a^{-1}(x) = f_a(x)$$ The graphs of functions $f_a$ and $f_a^{-1}$ are mirror reflections of each other in respect to the line $y=x$. By analyzing the graph of $f_a(x)$ we can identify several cases:
Case 1: $0<a<1$.
There is a single point $x_0\in(0,1)$ for which $$ f_a^{-1}(x_0) = f_a(x_0) = x_0$$ $x_0$ can be found using the Lambert W function: \begin{align} a^x = x &\Rightarrow xa^{-x} = 1 \Rightarrow \\ &\Rightarrow - x \ln a\cdot e^{-x\ln a } = -\ln a \Rightarrow \\ &\Rightarrow -x \ln a = W(-\ln a) \end{align} $$x_0 = \frac{1}{-\ln a}W_0(-\ln a) $$
However, it doesn't have to be the only solution of equation $f_a(x) = f_a^{-1}(x)$. Indeed, if we calculate $$ f_a'(x_0) - (f_a^{-1})'(x_0) = \ln a\cdot a^{x_0} - \frac{1}{x_0\ln a} = x_0\ln a - \frac{1}{x_0\ln a} = \frac{1}{W_0(-\ln a)} - W_0(-\ln a)$$ we see it can be of any sign; and if it is negative (that is $W_0(-\ln a) > 1$, equivalently $a<e^{-e}$), then with the facts that $$\lim_{x\rightarrow 0^+} (a^x - \log_a x) = -\infty$$ $$\lim_{x\rightarrow 1} (a^x - \log_a x) = a>0 $$ it means that there must be at least two additional roots, one in the interval $(0,x_0)$ and one in the interval $(x_0, 1)$.
To prove that for $0<a<e^{-e}$ there are exactly three solutions, and for $e^{-e}\le a<1$ there is exactly one solution, let us analyze the derivative of $a^x - \log_a x$:
\begin{align} & (a^x - \log_a x)' = 0 \Leftrightarrow \\ \Leftrightarrow & \quad \ln a\cdot a^{x} - \frac{1}{x\ln a} \Leftrightarrow \\ \Leftrightarrow & \quad x\ln a\cdot e^{x \ln a} = \frac{1}{\ln a} \Leftrightarrow \\ \Leftrightarrow & \quad x = \frac{1}{\ln a}W(\frac{1}{\ln a}) \end{align}
If $a> e^{-e}$ this equation doesn't have a solution, which leads to $(a^x - \log_a x)' > 0$ for all $x>0$ and $a^x - \log_a x$ is strictly increasing. That means it has only one root, and we have already found it, $x=x_0$.
The situation for border value $a= e^{-e}$ is similar. The derivative has only one root, which still means that $a^x - \log_a x$ is strictly increasing for $x>0$, and it has only one root $x=x_0 $.
If $a<e^{-e}$ the stuation is more complicated. Function $a^x - \log_a x$ is increasing on the interval $\big(0,\frac{1}{\ln a}W_0(\frac{1}{\ln a})\big)$, decreasing on $\big(\frac{1}{\ln a}W_0(\frac{1}{\ln a}),\frac{1}{\ln a}W_{-1}(\frac{1}{\ln a})\big)$ and again increasing on $\big(\frac{1}{\ln a}W_{-1}(\frac{1}{\ln a}),\infty\big) $. In each interval there can be at most one root, and we already know there are at least three roots for $a>e^{-e}$, so there are exactly 3 roots, one in each of these intervals. We know the exact location of only one, in the middle interval ($x=x_0$).
Case 2: $1<a<a_c$ where $a_c$ is the value of parameter $a$ for which the graph of function $a^x$ is tangent to the line $y=x$ (see Case 3).
In this case there are two points where $ f_a^{-1}(x) = f_a(x) = x$ given by $$ x_1 = \frac{1}{-\ln a}W_0(-\ln a) $$ $$ x_2 = \frac{1}{-\ln a}W_{-1}(-\ln a) $$
Case 3: $a=a_c$.
In this case there will be only solution, at the point where the graph $a^x$ is tangent to $y=x$. We can find both the value of $x_0$ and the value of $a_c$ by the conditions $$ \big(f_{a_c}(x_0) = x_0 \big) \land \big(f_{a_c}'(x_0) =1) $$ $$ \big({a_c}^{x_0} = x_0 \big) \land \big(\ln {a_c} \cdot {a_c}^{x_0} =1) $$ $$ \big(x_0 \ln {a_c} = 1\big) \land \big(\ln {a_c} \cdot e^{x_0 \ln {a_c}} =1)$$ $$ a_c = e^{1/e}, \qquad x_0 = e$$
Case 4: $a>a_c$.
In this case $f_a(x) > x$ and there are no solutions.
To sum up, there are \begin{array}{lll} \text{3 roots} & \text{for }0<a<e^{-e} & \text{one of them is }x=\frac{1}{-\ln a}W_0(-\ln a) \\ \text{1 root} & \text{for }e^{-e}\le x < 1 & x=\frac{1}{-\ln a}W_0(-\ln a) \\ \text{2 roots} & \text{for }1<a<e^{1/e} & x\in\{\frac{1}{-\ln a}W_0(-\ln a),\frac{1}{-\ln a}W_{-1}(-\ln a)\} \\ \text{1 root} & \text{for }x = e^{1/e} & x=e \\ \text{no roots} & \text{for }x > e^{1/e} & \end{array}
Actually finding $a(x)$ rather than $x(a)$ is easier: $$ e^{x\ln a} = \frac{\ln x}{\ln a }$$ $$ x\ln a \cdot e^{x\ln a} = x \ln x$$ $$ x\ln a = W(x\ln x)$$ $$ \ln a = \frac1x W(x\ln x)$$ which allows us to create a plot:
