Why are there $3$ solutions to $x^a = a^x$?

Question

Consider the equation, $x^a = a^x$.

If a is of the form $1.6 + 0.4k$, where $k$ is a natural number, then the equation has $3$ solutions. I checked them until $k = 1000$, but was unable to find a reason for this. Is there any plausible explanation for this? If yes, what is it? if not, why?

The possible duplicate question doesn't explain why there are 3 solutions for this particular form!

Thanks, John.

Answer 1

Taking the logarithm on both sides we get $$\frac{\ln(x)}{x}=\frac{\ln(a)}{a}$$ Now you can discover how many times cut $$\frac{\ln(a)}{a}$$ the graph of $$\frac{\ln(x)}{x}$$

Answer 2

The two positive solutions

On $x>0$, the value of $\frac{\ln x}{x}$ is monotonic increasing from $-\infty$ to $\frac{1}{e}$ and then monotonic decreasing, tending to $0$.

Therefore, for $a>1$ and $a\ne e$, the horizontal line $y=\frac{\ln a}{a}$ will be cut precisely twice by $\frac{\ln x}{x}$.

This will be obvious to you if you sketch the graph.

Negative solutions?

Instead of taking $x$ to the power $a=1.6 +0.4k$ you could consider not $x^a$ but $(x^2)^{0.8+0.2k}.$

You now have a fractional power of a positive number and so you can happily equate this expression with $a^x$. But please note that this is not the same as solving your original equation.

(This is, I believe, why you were convinced that you had solutions which others were saying did not exist.)

Answer 3

This is a partial answer.

Let us say that $a\neq 1$ and also $x\neq 1$, then we are interested in finding $x$ such that $$\frac{a}{\log a}=\frac{x}{\log x}.$$ If you consider the given domain, (by plotting a graph using a computer), we can see that there are precisely two different values for $x$ which satisfy this equation. This takes care of the two positive solutions. The third solution, if it exists should therefore be negative.

Fix $a>0$. Suppose there is a negative solution, call it $-y$ where $y>0$. Therefore we have that $$(-y)^a=\frac{1}{a^y}\Rightarrow (-1)^a(y)^a=\frac{1}{a^y}\Rightarrow \left(\cos(a\pi)+i\sin(a\pi)\right)(y)^a=\frac{1}{a^y}.$$

Since the right hand side is a real number (since $a>0$ and $y>0$), we should have that $\sin(a\pi)=0$, that is $a\in\mathbb{Z}$. Therefore whenever $a$ is non integral, there does not exist three solutions. Furthermore when $a$ is odd, we will have $$- y^a = a^{-y}$$ which is not possible because the LHS is a negative number and the RHS is a positive number (Again since $a>0$ and $y>0$). Therefore for a negative solution to exist, $a$ has to be even. In this case we are interested in solving, $y^a = a^{-y}$, where once again the $\log$ trick can be used.