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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

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    That's one possible approach. Ultimately, you will have to use the fact that $x\mapsto b^x$ is continuous, strictly increasing, and the intermediate value theorem.2010-12-21
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    Don't have IVT available yet (or anything about continuity).2010-12-21
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    If you don't have continuity, how do you define $b^x$ for arbitrary real $x$? I often see it defined first for $x\in\mathbb{N}$, then for $b > 0$ and $x\in \mathbb{Q}$ by taking roots, and then assert that $b^x$ is defined for $x\in\mathbb{R}$ "by continuity"...2010-12-21
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    @Willie: You can define it as the supremum of the set of $b^r$ where $r$ is rational and less than $x$. @Trevor: Could you please cite the source of your problem if this comes from a textbook? The context would make it clearer what sort of answer you're looking for. If you don't have a source, then please elaborate in your question on how things are defined.2010-12-21
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    This is from Chapter 1 from Baby Rudin. He goes through an outline of the proof. But I am just trying to do it from scratch (what he does is tricky).2010-12-21

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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}

(d) If $b^w

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}

(e) If $b^w>y$, then $b^{w-(1/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}

(f) Let $A = \{w : b^w

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), $b^{x+(1/n)} x$ 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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    Second half of the first inequality, n/b(b-1), do you mean (n/b)*(b-1) or. n/[b*(b-1)]?2015-10-05
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    I think it's the first of the two possibilities. It's still ambiguous.2015-10-05
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    Your proof of (g) is problematic. Though we know $x=\sup A$ leads to $b^x=y$, $b^z=y$ doesn't necessarily lead to $z=\sup A$2017-09-06
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    @Hank not really.2017-09-06
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    Real exponentiation is injective. If $b^{\mathrm{sup} A} = y = b^z$, then $z=\mathrm{sup} A$.2017-09-06
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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 $\{r\in\mathbb{R}\mid b^r\leq y\}$).