# Prove the existence of logarithms

1. May 21, 2008

### tronter

Fix $b >1, \ y >0$, and prove that there is a unique real $x$ such that $b^{x} = y$.

Here is the outline:

(a) For any positive integer $n$, $b^{n}-1 \geq n(b-1)$. Why do we do this?

(b) So $b-1 > n(b^{1/n}-1)$.

(c) If $t>1$ and $n > (b-1)/(t-1)$ then $b^{1/n} < t$.

etc..

Is this the correct process?

2. May 21, 2008

### rootX

my approach:
draw graph of b^x .. and use some theorm (I forgot it's name) .. I think intermediate value theorm