Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 1, Problem 7

Problem

Fix $b>1$, $y>0$, and prove that there is a unique real $x$ 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 $n$; to see this, apply part (c) with $t=y \cdot b^{-w}$.

(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.

Answer

(a) Note that

$$b^n - 1 =(b-1)(1+b+b^2 + \cdots + b^{n-1}).$$

Since $b>1$,

$$1+b+b^2+\cdots+b^{n-1} \geq n.$$

Hence

$$b^n - 1 = (b-1)(1+b+b^2 + \cdots + b^{n-1}) \geq n(b-1).$$

(b) It is enough to show that $b^{1/n} > 1$. If $b^{1/n} \leq 1$, then $b = (b^{1/n})^n \leq 1^n = 1$, which is contradiction.

(c) Since $t>1$,

$$n(t-1) > (b-1) \geq n(b^{1/n} - 1).$$

$b^{1/n}<t$ is obtained by dividing both sides by $n$ and adding 1.

(d) For $t:=yb^{-w}>1$, we can take an integer $n$ such that $n>(b-1)/(t-1)$ by archimedean property. Then, (c) implies

$$b^{1/n} < t = yb^{-w}\;\;\Longrightarrow\;\; b^{w+1/n}<y.$$

(e) For $t:=y^{-1}b^w>1$, we can take an integer $n$ such that $n>(b-1)/(t-1)$ by archimedean property. Then, (c) implies

$$b^{1/n} < t = y^{-1}b^{w} \;\;\Longrightarrow\;\; y < b^{w-1/n}.$$

(f) If $b^x < y$ then there exists an integer $n$ such that $b^{x+1/n} < y$, which means $x < x+1/n \in A$. Therefore, $x$ is not an upper bound of $A$.

If $b^x>y$ then there exists an integer $n$ such that $b^{x-1/n} > y$, which means

$$b^{x-1/n} > y > b^w \;\;\Longrightarrow\;\; b^{w-x}<b^{-1/n}\;\;\text{for all $w \in A$}.$$

Since $b^{w-x} = \sup\{b^t | t \leq w-x, t \in \mathbb{Q}\}$ by Problem 6, $w-x < -1/n$ for all $w \in A$, which implies $x-1/n$ is an upper bound for $A$. Therefore, $x$ cannot be a least upper bound of $A$.

Since both cases above lead to contradiction, $b^x=y$.

(g) Assume that there exists $z \in \mathbb{R}$ such that $y=b^z$. Without loss of generality, it is enough to show that $x>z$ leads to the contradiction. Suppose $x>z$, and then there exists an integer $n$ such that $x-z>1/n$. Then $1 = yy^{-1} = b^xb^{-z} = b^{x-z} \geq b^{1/n} > 1$, which is contradiction.

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