Existence of Real Logarithm

Theorem
Let $b, y \in \R: b > 1, y > 1$.

Then there exists a unique real $x \in \R$ such that $b^x = y$.

This $x$ is called the logarithm of $y$ to the base $b$.

Also see the definition of a (general) logarithm.

Proof
(a) First note that for any positive integer $n$, $b^n - 1 \ge n \left({b-1}\right)$.

This is true as each of $b^{n-1},b^{n-2}, \ldots, b$ are greater than $1$, so summing and then applying the formula of the sum of a finite geometric series proves the claim.

(b) The second step involves proving that $b-1 \ge n \left({b^{1/n} - 1}\right)$.

This follows, because as $b^{1/n} > 1$, by (a), $\left({b^{1/n}}\right)^n - 1 \ge n \left({b^{1/n} - 1}\right)$.

(c) The next step is showing that if $t > 1$ and $n > \dfrac {b - 1} {t - 1}$ then $b^{1/n} < t$.

This follows from $b^{1/n} = \left({b^{1/n} - 1}\right) + 1 \le \dfrac{b - 1} {n} + 1 < t$ if we use the claim established in (b).

(d) The next step is showing that if $w$ is such that $b^w < y$ then $b^{w + (1/n)} < y$ for sufficiently large $n$.

To see this note that $1 < b^{-w}y = t$ (say).

Choose $\displaystyle n > \frac {b - 1} {t - 1}$, then by (c), $b^{1/n} < b^{-w} y$ or $b^{w + (1/n)} < y$ for sufficiently large $n$.

(e) Similar to (d) it can be now noted that if $b^w > y$, then $b^{w-(1/n)} > y$ for sufficiently large $n$.

(f) The following claim can now be established:

Let $A$ be the set of all $w$ such that $b^w < y$.

Then $x = \sup A$ satisfies $b^x = y$.

Proof:

From (a), $b^n \ge n \left({b - 1}\right) + 1$ for all $n$.

For which each $z \in \R$ choose an $n$ so that $n \left({b - 1}\right) > z - 1$, or $n \left({b - 1}\right) + 1 > z$.

Hence for all $z$ we have an $n$ such that $b^n \ge \left({b - 1}\right) + 1 > z$.

Hence the set $\left\{{b^n : n \in \N}\right\}$ is unbounded.

Now consider the function $f: \R \to \R$ defined by $f(x) = b^x$.

Clearly $f$ is a strictly increasing function. This follows from the definition of $b^x$.

Define $A = \left\{{w: b^w < y}\right\}$.

The set $\left\{{b^n: n \in \N}\right\}$ being unbounded gaurantees the existence of a $n$ such that $b^n > y$.

Thus $n$ is an upper bound for $A$.

Let $x = \sup A$.

Suppose $b^x < y$.

By (d), for sufficiently large $n$, $b^{x + (1/n)} < y$, i.e. $x + 1/n \in A$.

But this is impossible as $x = \sup A$.

So $b^x < y$ is not possible.

Suppose $b^x > y$.

By (e), for sufficiently large $n$, $b^{x - (1/n)} > y$, i.e. $x - 1/n \notin A$.

Now $x - 1/n$ cannot possibly be the supremum of $A$.

So there is a $w \in A$ such that $x - 1/n < w \le x$.

But then as $f$ was strictly increasing, $b^{x - 1/n} < b^w < y$.

This is a contradiction, as $b^{x - (1/n)} > y$.

So $b^x > y$ is not possible.

(g) The fact that this $x$ is unique follows from the fact the function $f$ described in (f) is increasing and hence, by Strictly Monotone Mapping is Injective, is 1-1.