Euler's number inherently in $\mathbf{x^a=a^x}$ and $\mathbf{x.a=a^x}$?

Question

I was solving some equations graphically, when I came across an identity of Euler's number in two cases:

Case 1: $\mathbf{a.x = a^x}$

Ever wondered how for two numbers $\mathbf{\left\{{ a,x,y} \right \}\in \mathbb{R_{> 0}} :\space a.x = a^x= y}$?

For example natural numbers: $\mathbf{2.2=2^2=4}$

1. What is the general formula for $\mathbf{x}$, given that $\mathbf{x.a=a^x}$?

It has multiple solutions. One is always $\mathbf{x=1}$ And interestingly, if $\mathbf{x_1=x_2: \space a=e}$. In this case, actually: $\mathbf{x_1=x_2=1}$. So as $\mathbf{a\to e\implies x_1=x_2\to1}$.

$\mathbf{e}$ "came out of nowhere". I am sure the analytical solution would introduce something like Lambert W function which would explain the presence of $\mathbf{e}$.

Case 2: $\mathbf{x^a = a^x}$

Or second case when: $\mathbf{\left\{{ a,x,y} \right \}\in \mathbb{R_{> 0}} :\space x^a = a^x= y}$?

For example $\mathbf{2^2=2^2=4}$ or how $\mathbf{3^{2.478...}\approx2.478...^3\approx15.216}...$

2. What is the general formula for $\mathbf{x}$, given that $\mathbf{x^a=a^x}$?

It has multiple solutions. And interestingly, as $\mathbf{x\to e\implies x_1\to x_2}$.

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What are the two formulas for $x$ and how to obtain them?

Solution attempt

How to solve for $x$?

I tried this:

$\mathbf{log_ax^a = log_aa^x\\a.log_ax=x\\a.x^{-1}log_ax=1\\a.log_ax^{x^{-1}}=1}\\$

Lambert W function comes into mind. But how?

Answer 1

It is the answer that provides the solution to your second equation:

$$ax=a^x$$

$$\frac{a^x}{x}=a$$

Let, $a^x=t$ and $x\ln a=\ln t$, then we get

$$\frac{t}{\frac {\ln t}{\ln a}}=a$$

$$\frac {t}{\ln t}=\frac {a}{\ln a}$$

$$\frac {\ln t}{t}=\frac {\ln a}{a}$$

Based on the first answer, we have

$$t=e^{-W\left(-\frac{\ln a}{a}\right)}$$

$$x=\frac {\ln t}{\ln a}=\frac {-W\left(-\frac{\ln a}{a}\right)}{\ln a}.$$

Answer 2

This answer deals with the solution to the first equation:

$$x^a=a^x$$

$$a\ln x=x\ln a$$

$$\frac{\ln x}{x}=\frac{\ln a}{a}$$

Let, $x=t^{-1}$

$$t\ln t=-\frac{\ln a}{a}$$

$$W\left(\ln te^{\ln t}\right)=W\left(-\frac{\ln a}{a}\right)$$

$$\ln t=W\left(-\frac{\ln a}{a}\right)$$

$$\ln x=-W\left(-\frac{\ln a}{a}\right)$$

$$x=e^{-W\left(-\frac{\ln a}{a}\right)}.$$