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 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 $$n > \frac {b - 1} {t - 1}$$ then $$b^{1/n} < t$$.

This follows from $$b^{1/n} = \left({b^{1/n} - 1}\right) + 1 \le \frac{b - 1} {n} + 1 < t$$ 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 $$n > \frac {b - 1} {t - 1}$$, 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.