Here's problem 7 in the exercises following Chap. 1 in Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Fix $b > 1$, $y > 0$, and prove that there is a unique real number $x$ such that $b^x = y$, by completing the following outline. (This $x$ is called the logarithm of $y$ to the base $b$. )
(a) For any positive integer $n$, $\ $ $b^n-1 \geq n (b-1)$.
(b) Hence $b-1 \geq n (b^{1/n} - 1)$.
(c) If $t > 1$ and $n > (b-1)/(t-1)$, then $b^{1/n} < t$.
(d) If $w$ is such that $b^w < y$, then $b^{w+1/n} < y$ for sufficiently large $n$; ...
(e) If $b^w > y$, then $b^{w-1/n} > y$ for sufficiently large $n$.
(f) Let $A$ be the set of all $w$ such that $b^w < y$, and show that $x = \sup A$ satisfies $b^x = y$.
(g) Prove that this $x$ is unique.
Now (a) can be proved using induction on $n$, and (b) can be obtained by putting $b^{1/n}$ for $b$ in (a) since $b > 1$ implies $b^{1/n} > 1$ also.
(c) If $t > 1$ and $n > (b-1)/(t-1)$, then we have $$n (t-1) > b-1 \geq n(b^{1/n} -1) \ \ \ \mbox{ using (b)}. $$ So $$t-1 > b^{1/n} -1,$$ from which the desired result follows.
(d) If $w$ is such that $b^w < y$, then $y b^{-w} > 1$ and we put $t \colon= y b^{-w}$ in (c) to obtain $b^{1/n} < y b^{-w}$ or $b^{w+1/n} < y$ for a sufficiently large $n$ (i.e., whenever $n > (b-1)/(yb^{-w}-1)$).
(e) If $w$ is such that $b^w > y$, then $b^w y^{-1} > 1$, so as in (c), if we choose a positive integer $n$ such that $n > (b-1)/(b^w y^{-1} -1)$, then we have $b^{1/n} < b^w y^{-1}$ and so $b^{w-1/n} >y$.
Now let $$ S \colon= \{ \ b^n \ \colon \ n \in \mathbb{N} \ \}.$$ We show that the set $S$ is not bounded above in $\mathbb{R}$. Let's assume the contrary. Let $\alpha \colon = \sup S$. Then $\alpha \geq b^n$ for all $n \in \mathbb{N}$; so in particular, $\alpha > 1$. Now as $\alpha/b < \alpha$, so $\alpha/b$ is not an upper bound for $S$. Thus there exists a $k \in \mathbb{N}$ such that $\alpha/b < b^k$. But then $\alpha < b^{k+1}$ and $k+1 \in \mathbb{N}$ also. Hence a contradiction.
Therefore, for any real number $\alpha$, there is a natural number $n$ such that $b^n > \alpha$.
We can also show that, for any integers $m$ and $n$, $b^m < b^n$ if and only if $m< n$.
(f) We first need to show that the set $A$ is non-empty and bounded above.
If $y> 1$, then $0 \in A$, for $b^0 =1 < y$. For $y=1$, $-1\in A$, for $b^{-1} = 1/b < 1$. For $0< y< 1$, we have $1/y > 1$. So there exists a natural number $n$ such that $b^n > 1/y$. So for this $n$, we have $b^{-n} < y$ and so $n \in A$. Thus the set $A$ is non-empty.
Now we show that the set $A$ is bounded from above.
We can prove the following statement:
For any rational numbers $p$, $q$, we have $b^p < b^q$ if and only if $p < q$.
Since there is a positive integer $n$ such that $b^n \geq y$, there is a least positive integer $n_0$ with this property. Now if $w$ is a real number such that $w > n_0$, then we can find a rational number $r$ such that $n_0 < r < w$, and for this $r$ we have $b^{n_0} < b^r$ since $n_0$ is also rational. Therefore, using the discussion in Prob. 6, Chap. 1 in Baby Rudin, we can write $$ \begin{align} b^w &= \sup B(w) = \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq w \ \} \\ &\geq \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq r \ \} = \sup B(r) \\ &= b^r \\ &> b^{n_0} \\ &\geq y.\end{align} $$ That is, $b^w > y$ if $w > n_0$; so $w \leq n_0$ for all $w \in A$, and the set $A$ is bounded above.
Let $x \colon= \sup A$. We show that $b^x = y$.
If $b^x < y$, then using part (d) above, we can find a positive integer $n$ such that $b^{x+1/n} < y$ and so $x+1/n \in A$ for this choice of $n$. But $x+1/n > x$. So we have a contradiction in view of our choice of $x$ as an upper bound for the set $A$. Hence $b^x \not< y$.
If $b^x > y$, then using part (e) above, we can find a positive integer $n$ such that $b^{x-1/n} > y$ and so if $w > x-1/n$, then $b^w > y$, which implies that $w \leq x-1/n$ for all $w \in A$, showing that $x-1/n$ is also an upper bound for $A$. But $x-1/n < x$. So we have a contradiction in view of our choice of $x$ as the least upper bound for $A$. Hence $b^x \not>y$.
Therefore, $b^x = y$.
(g) Finally, we show that this $x$ is unique. Suppose that, for some $z > x$, we also have $b^z = y = b^x$.
Since $z > x$, we can choos rational numbers $p$ and $q$ such that $x < p < q < z$, and therefore, we have $$ \begin{align} b^z &= \sup B(z) = \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq z \ \} \\ &\geq b^q \\ &> b^p \\ &\geq \sup \{ \ b^t \ \colon \ t \in \mathbb{Q}, \ t \leq x \ \} = \sup B(x) \\ &= b^x, \end{align} $$ and hence $b^z > b^x$, which contradicts our choice of $z$.
So our real number $x$ such that $b^x = y$ is indeed unique.
Is this proof correct? Or, are there any holes in it?