Solve the system of equations $x^y=y^x$

Question

Solve the system of equations
$$ x^y=y^x \\ a^x=b^y $$

I could not solve this despite many tries

Answer 1

I assume you are looking for solutions where $x,y,a,b$ are positive, none equal to $1$. Furthermore, that $a\neq b$. [If $a=b$, then any pair $y=x$ is a solution.] The equations can be rearranged to

$$\left\{\begin{aligned} y\ln(x)&=x\ln(y)\\ y&=x\frac{\ln(a)}{\ln(b)}=cx \end{aligned}\right.$$ where $c=\frac{\ln(a)}{\ln(b)}\neq1$. Substituting $y$ from the second into the first: $$ \begin{align} &&cx\ln(x)&=x\ln(cx)\\ \implies&& c\ln(x)&=\ln(cx)\\ \implies&& x^c&=cx\\ \implies&& x^{c-1}&=c\\ \end{align} $$ If $c$ is negative (that is, if $a,b$ are on opposite sides of $1$), there are no solutions. Otherwise, if $a,b$ are on the same side of $1$, then $x=c^{1/(c-1)}$ and:

$$\begin{align} x&=\left(\frac{\ln(a)}{\ln(b)}\right)^{1\left/\left(\frac{\ln(a)}{\ln(b)}-1\right)\right.}=\sqrt[\log_b(a)-1]{\log_b(a)}\\ y&=cx=\left(\frac{\ln(a)}{\ln(b)}\right)^{1\left/\left(1-\frac{\ln(b)}{\ln(a)}\right)\right.}=\sqrt[1-\log_a(b)]{\log_b(a)} \end{align}$$