1. ## uniqueness of log

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?

2. In other words the strategy is that we need to consider a set....show that $\displaystyle x$ is the supremum of that set...and then show uniqueness?

3. See what is the point of showing that is is strictly increasing? I see. Because you can show uniqueness by exploiting monotonicity.

4. Let us consider the slightly different problem of showing that there is a unique $\displaystyle b$. So we fix $\displaystyle x, y>0$. Then uniqueness follows because $\displaystyle 0 < y_1 < y_2 \implies y_{1}^{n} < y_{2}^{n}$. And to show existence we consider the set of $\displaystyle t$'s such that $\displaystyle t^n < x$.

So we use this same set in our problem. But we are trying to use this strategy: $\displaystyle \text{monotonic increasing} \implies \text{uniqueness}$.