Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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?

share|improve this question
    
What do you want solve the inequality for? A simplified expression in $x,y$ or is it something else?? –  Ramana Venkata Dec 20 '11 at 7:52
    
@Ramana: It appears that Charles wants to show that the inequality in the first line is equivalent to the simpler inequality $x\ge -y$. –  Brian M. Scott Dec 20 '11 at 9:07

1 Answer 1

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?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.