# Thread: Need help proving a relationship

1. ## Need help proving a relationship

I have to prove that

a^log n = n^log a

an example would be
2^log n = n^log2 = n

2. Originally Posted by darken4life
I have to prove that

a^log n = n^log a

an example would be
2^log n = n^log2 = n
That isn't quite a true statement. It only works for a specific base:
$\displaystyle 2^{log_2(n)} = n^{log_2(2)} = n$

Prove: $\displaystyle a^{log_a(n)} = n^{log_a(a)}$

This isn't so much a proof as a definition: the logarithm and exponentiation operations are functional inverses.

What you need to do is show that each of these are equivalent to the same number.

We know that $\displaystyle n^{log_a(a)} = n^1 = n$

So what is $\displaystyle a^{log_a(n)}$? Well, call it x.

Then
$\displaystyle x = a^{log_a(n)}$

$\displaystyle log_a(x) = log_a \left ( a^ {log_a(n)} \right )$

$\displaystyle log_a(x) = log_a(n)$

Since log is a 1 to 1 function, we must have x = n. Thus
$\displaystyle a^{log_a(n)} = n = n^{log_a(a)}$

-Dan