If $x^y = y^x$ $(x,y \in R, x,y>0,x\neq 0 )$ and $x^p = y^q$ $(p,q \in R/\{0\}, p \neq q)$, then product $xy$ is equal to?

Question

Solution for this one is $({\frac{p}{q}})^{\frac {p+q}{p-q}}$ , but I do not understand how I am supposed to get here, I guess something with logarithms but not sure what?

Answer 1

So $x=y^{q/p}$. Thus you have $y^{qy/p}=y^{y^{q/p}}$. As $y>0$, we have $qy/p=y^{q/p}$ i.e. $y^{(q-p)/p}=q/p$ i.e. $y=(q/p)^{p/(q-p)}$.

Hence $x=(q/p)^{q/(q-p)}$.

So $xy=(q/p)^{(p+q)/(q-p)}=(p/q)^{(p+q)/(p-q)}$

Answer 2

HINT:

Let $x^p=y^q=z^{pq}$

$\implies$ one of the values of $x$ is $=z^q$ and $y=z^p$

$x^y=y^x\implies(z^p)^{z^q}=(z^q)^{z^p}$

$\implies pz^q=qz^p\implies z^{q-p}=\dfrac qp$

$\implies$ one of the values of $z$ is $\left(\dfrac qp\right)^{1/(q-p)}=\left(\dfrac pq\right)^{1/(p-q)}$

Now $xy=z^{p+q}=\cdots$