Simplifying $x \gt x^{\ln 2/\ln 3}$ algebraically to $x \gt 1$

I simplified an inequality to the following point $x > x^{\ln 2/\ln 3}$, I know the answer is $x > 1$.

Of course one could explain that $y = x$ exceeds $y = x^n$ where $0<n<1$ at $x>1$
given the behaviour of that set of functions.

But is there a way that this can be shown algebraically?
For example, by converting to logarithmic form,
simplifying then reconverting to exponential form?

Thanks for the help.

Notice that the inequality is equivalent to $$ x - x^{\log_3(2)}> 0 \color{Blue}{\implies} x\left(1-x^{\log_3(2)-1} \right)>0 $$ In the above you have two factor being multiplied which give a positive number, so both factors must be positive or both factors must be negative. Then analyze each case and this will lead to your answer.

The RHS of the inequality makes no sense if $x<0$, so assume $x\ge0$ in what follows: $$x>x^{\ln2/\ln3}=x^{\log_32}$$ $$1>x^{\log_32/3}$$ Now $\log_3\frac23<0$ and the graph of $x^a$ for $a<0$ is monotone decreasing with $1^a=1$, so we get $x>1$.

There is a way to solve this where the only property used about the natural logarithmic function is that it is strictly increasing:

Let $a=\frac{\ln 2}{\ln 3}$. Then $a < 1$. Thus we are left with finding the set of $x > 1$ that satisfy $x < x^a$ for some $a <1$, or equivalently, $x^{b}<1$ for $1-a = b > 0$.

As $f(y)=y^b$ is increasing in $y$ and $1^b =1$ it follows that all $x>1$ satisfy $x^b <1$, and thus the original inequality $x<x^a$, where $a=\frac{\ln 2}{\ln 3}$.