# Help with Inequality with Exponents question

1. Mar 22, 2010

### magic_castle32

How do you prove the following?

$$7^{\sqrt{5}} > 5^{\sqrt{7}}$$

Without calculators or working out $${\sqrt{5}}$$, $${\sqrt{7}}$$, of course.

2. Mar 22, 2010

### l'Hôpital

Well, consider both numbers and raise them both to the power of square root of 5.

So we have
$$7^5 \ and \ 5^{\sqrt{35}}$$

Now, consider the following:

$$6 = \sqrt{36} > \sqrt{35}$$

So we can just compare 7^5 and 5^6. I'm sure you can calculate those by hand.