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

So to show uniqueness we need to consider the following: if $\displaystyle b^{x} = y $ and $\displaystyle b^{l} = y $ then $\displaystyle y = l $. But first we want to show that $\displaystyle b^{x} $ is monotonically increasing.

So for any positive integer $\displaystyle n $, $\displaystyle b^{n}-1 \geq n(b-1) $. We can show this by factoring the LHS as $\displaystyle (b-1)(b^{n-1}+b^{n-2}+b^{n-3} + \ldots + b^{2} + b+1) \geq (b-1)n $. So $\displaystyle b-1 \leq \frac{b^{n}-1}{n} $. Or $\displaystyle b-1 \geq n(b^{1/n}-1) $. From here what do we do?