
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(b1) $. We can show this by factoring the LHS as $\displaystyle (b1)(b^{n1}+b^{n2}+b^{n3} + \ldots + b^{2} + b+1) \geq (b1)n $. So $\displaystyle b1 \leq \frac{b^{n}1}{n} $. Or $\displaystyle b1 \geq n(b^{1/n}1) $. From here what do we do?

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?

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

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} $.