Proving that if $a>1$ and $x>y$ then $a^x>a^y$

Question

I got this assignment for homework and I can't find this anywhere around the web.

Prove that if $a>1$ and $x>y$ then $a^x>a^y$.

I started the assignment but I'm not sure it's enough:

$n>0$

$x=y+n$

$a^x=a\times a\times a\times\dotsb\times a$ ($y+n$ times)

$a^y=a\times a\times\times\dotsb\times a$ ($y$ times)

$y+n>y$ and $a>0$ then $a^x>a^y$.

Please help me.

Hi, please show us what you have done? – meta_warrior Nov 15 '13 at 16:38 — meta_warrior, Nov 15 '13 at 16:38
Anyway the hint is to use $x-y>0$ – meta_warrior Nov 15 '13 at 16:42 — meta_warrior, Nov 15 '13 at 16:42

score 2 · Answer 1 · answered Nov 15 '13 at 16:47

2

Hint: $x-y>0,$ so since $a>1,$ what can you say about $a^{x-y}$? What can you conclude from there?

(The problem with your approach is that it assumes that $x,y$ differ by a positive integer.)

answered Nov 15 '13 at 16:47

Cameron Buie

score 0 · Answer 2 · edited Nov 16 '13 at 03:04

0

Since $x>y$

Multiply $\ln(a)$ on both sides.

The inequality symbol remains the same since $\ln (a) > 0 $ as $a>1$

Therefore $x \ln(a) > y \ln(a)$

Hence $\ln(a^x) > \ln(a^y)$

Since $\ln$ is a logarithm with base "$e$" ($e>1$ )

Therefore $a^x > a^y$ .

edited Nov 16 '13 at 03:04

Michael Hardy

answered Nov 15 '13 at 16:58

Mahesh

2 Answers2