Logarithm and Uniqueness Fix $b>1, y>0$ and prove that there is a unique real $x$ such that $b^x = y$. So for uniqueness, $x_1 < x_2 \Rightarrow b^{x_1} < b^{x_2}$. Then consider the set $S = \{b^t: t \leq x \}$ and work from there? 
Source: Chapter 1, Problem 7, Principles of Mathematical Analysis by Rudin
 A: Actually, the proof is quite easy following his outline:
(a) Lemma: For any positive integer $n$, $b^n -1 \geq n(b-1)$
Proof: Suppose this holds for $n$. Then $b^n-1\geq n(b-1)\geq n/b(b-1)$, since $b>1$
Then $b^{n+1} \geq n(b-1)+b$, and so $b^{n+1}-1 \geq n(b-1)+(b-1)$, therefore $b^{n+1}-1\geq (n+1)(b-1)$. Since $b^n -1 \geq n(b-1)$ holds for $n=1$, by induction it holds for any $n$.
(b) Hence, $b-1 \geq n(b^{1/n}-1)$ if we apply the lemma setting $b' =b^n$ ($b'>1$ since $b>1$).
(c) If $t>1$ and $n>\frac{b-1}{t-1}$, then $b^{1/n} < t$.
Proof: Since $n>\frac{b-1}{t-1}$, then $n(t-1)>(b-1)\geq n(b^{1/n}-1)$. It follows that $b^{1/n} <t$.
(d) If $b^w<y$, then $b^{w+(1/n)}<y$ for sufficiently large $n$.
If we set $t=y/b^w$, we see that $t>1$ and we can choose a sufficiently large $n$ such that $n >\frac{b-1}{t-1}$. Because of (c) we know that $b^{1/n}<t$. Multiply both sides by $b^w$ to get $b^{w+1/n}<y$, which is what we wanted to prove.
(e) If $b^w>y$, then $b^{w-(1/n)}<y$ for sufficiently large $n$.
If we set $t=b^w/y$, we see that $t>1$ and we can choose a sufficiently large $n$ such that $n >\frac{b-1}{t-1}$. Because of (c) we know that $b^{1/n}<t$. Then $1<\frac{t}{b^{1/n}}$, so $1<\frac{b^w}{yb^{1/n}}$ and therefore $y<b^{w-(1/n)}$.
(f) Let $A = \{w : b^w<y\}$, then if $x=\sup A$, $b^x = y$.
Since $b>1, y>0$, choosing a sufficiently small $w$ will get us arbitrarily close to $0$, so $A$ is non-empty. Since $b^w>y$ for a sufficiently large $w$, $A$ is bounded above, and because of $A$ being a subset of $\mathbb{R}$ then $\exists\sup A$.
Fix $x = \sup A$. Suppose $b^x < y$, then because of (d), $b^{x+(1/n)}<y$ for some $n$. But then $x+1/n > x$ and $x+1/n \in A$, which is a contradiction.
Suppose $b^x > y$, then because of (e), $b^{x-(1/n)}>y$ for some $n$. But then $x-(1/n)$ is an upper bound of $A$ smaller than $x$, which is a contradiction.
Therefore if $x=\sup A$, then $b^x=y$
(g) The uniqueness of $x$ follows from the fact that there can't be two different least upper bounds for the same set $A$.
Feedback is appreciated, thanks.
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A: Yes, if you prove that $x\mapsto b^x$ is strictly increasing, that will show that there is at most one solution.
However, your proposal with $S$ does not really work because what you are looking for is $x$; since you do not know ahead of time what $x$ is, you cannot define $S$ in the first place!
Instead, consider the set $S = \{ r\in\mathbb{Q} \mid b^r\leq y\}$ (or $\{r\in\mathbb{R}\mid b^r\leq y\}$). 
