Is it coincidential that $\left(\frac{px^a}{qx^b}\right)^x$ always intersects $\frac{p^x\cdot x^a\cdot x}{q^x\cdot x^b\cdot x}$ once

Question

Take for example, $x^x=(\frac {x}{1})^x $, then it will have one point of intersection with $\frac {x^2}{x}=x $

Or $\left(\frac{2x^2}{3x^3}\right)^x\implies\frac{2^x\cdot x^3}{3^x\cdot x^4}$

Also works with: $\left(\frac{x^2}{x^4}\right)^ {x^2}\implies\frac{x^4}{x^6}$

Shorter rule: One point of intersection $r$ such that $\left(\frac{px^a}{qx^b}\right)^x=\frac{p^x\cdot x^{a+1}}{q^x\cdot x^{b+1}}$

Note: One point of intersection from the interval $x>0$.

Answer 1

You are asking for solutions of $$ \left(\frac{px^a}{qx^b}\right)^x=\frac{p^xx^{a+1}}{q^xx^{b+1}} $$ That is,$$ \frac{p^xx^{ax}}{q^xx^{bx}}=\frac{p^xx^{a+1}}{q^xx^{b+1}}$$ Now we can cancel $p^x/q^x$ and then subtract exponents to get $$ x^{(a-b)x}= x^{a-b}$$ Now the exponents must be equal, so either $a=b$ or $x=1$.

In the trivial case that $a=b$ it's true for every $x>0$. Otherwise, $x=1$ is the only solution.