# What is an easy way to prove the equality $r > s > 0$ implies $x^r > x^s$?

I have been using simple inequalities of fractional powers on a positive interval and keep abusing the inequality for $x>1$. I was just wondering if there is a nice way to prove the inequality in a couple line:

Let $x \in [1,\infty)$ and $r,s \in \mathbb{R}$

What is an easy way to prove the equality $r > s > 0$ implies $x^r > x^s$?

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The question should be changed to make the inequality true. If $x=1/2$ and $r>s>0$, then $x^r<x^s$. –  André Nicolas Oct 4 '11 at 5:03
Most of the needed change was made. But we don't want $x=1$. –  André Nicolas Oct 4 '11 at 5:13
$r\gt s$ implies $r\log x\gt s\log x$ which is $\log(x^r)\gt\log(x^s)$ which implies $x^r\gt x^s$ if you are willing to accept that the logarithm function is monotone increasing. The first implication in the chain relies on $\log x\gt0$ which is in line with Andre's comment and highlights the need for further restrictions on $x$.
If you accept that $x^y\gt 1$ for $x\gt 1$ and $y \gt 0$, then $x^r=x^{r-s}x^s \gt x^s$ for $x\gt 1$ and $r \gt s$.