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$(\log_23)^x-(\log_53)^x\geq(\log_23)^{-y}-(\log_53)^{-y}$

I guess the function $f(x)=(\log_23)^x-(\log_53)^x$ monotonically increasing, so I get the answer $x\geq-y$,but how to prove it not using calculus?

Or how to proof $f(x)=a^x-b^x(a>b)$ is monotonically increasing?

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  • $\begingroup$ What do you want solve the inequality for? A simplified expression in $x,y$ or is it something else?? $\endgroup$ – Ramana Venkata Dec 20 '11 at 7:52
  • $\begingroup$ @Ramana: It appears that Charles wants to show that the inequality in the first line is equivalent to the simpler inequality $x\ge -y$. $\endgroup$ – Brian M. Scott Dec 20 '11 at 9:07
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For convenience let $a=\log_23$ and $b=\log_53$. Note that $a>1$ and $b<1$. Thus, $a^x$ increases as $x$ increases, while $b^x$ decreases, and therefore $a^x-b^x$ does what?

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