An inequality regarding power of two real numbers

Assume we have two real numbers $a,b\in \mathbb{R}$ and we know $a,b> 1$. We do not know anything about their relative magnitude, i.e. all of these $a\le b, a\ge b, a=b$ are possible. However we do know that $a^{x_1} = b^{x_2}$ and $0<x_2 < x_1$.

My question is: can one conclude that $ax_1\le bx_2$?

I know that $a\le b$ can be concluded (because $x_1\ln a = x_2 \ln b$ and $\ln x$ is an increasing function), but I don't know about that claim or how to approach the proof. Is this comparison even correct/possible?

Thank you.

The question was edited after I posted this answer.

$a=1, b=2, x_1=1, x_2=0$ gives a counter-example

Hint for the edited version: Let $a=2,b=3$ so that $b^{a} >a^{b}$. Let $x_1=2$ and define $x_2$ by $a^{x_1}=b^{x_2}$ (which means $x_2 =x_1 \frac {\ln a } {\ln b}$). Now $ax_1>bx_2$ because $a>b \frac {\ln a } {\ln b}$.