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

Question

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.

Answer 1

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.

Answer 2

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 $x>1$.

Answer 3

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 $a=\frac{\ln 2}{\ln 3}$.