Different definitions of logarithm (Part 6)

There are two apparently different definitions of a logarithm that appear in the secondary mathematics curriculum:

1. From Algebra II and Precalculus: If $a > 0$ and $a \ne 1$, then $f(x) = \log_a x$ is the inverse function of $g(x) = a^x$.
2. From Calculus: for $x > 0$, we define $\ln x = \displaystyle \int_1^x \frac{1}{t} dt$.

The connection between these two apparently different ideas begins with the following theorem.

Theorem. Let $a \in \mathbb{R}^+ \setminus \{1\}$. Suppose that $f: \mathbb{R}^+ \rightarrow \mathbb{R}$ has the following four properties:

1. $f(1) = 0$
2. $f(a) = 1$
3. $f(xy) = f(x) + f(y)$ for all $x, y \in \mathbb{R}^+$
4. $f$ is continuous

Then $f(x) = \log_a x$ for all $x \in \mathbb{R}^+$.

Note. To prove this theorem, I will show that $f(a^x) = x$, thus proving that $f$ is the inverse of $g(x) = a^x$.

The proof of these theorem divides into four cases:

1. Positive integers: $x = m \in \mathbb{Z}^+$
2. Positive rational numbers: $x = \frac{m}{n}$, where $m,n \in \mathbb{Z}^+$
3. Negative rational numbers: $x \in \mathbb{Q}^-$
4. Real (possibly irrational) numbers: $x \in \mathbb{R}$

In today’s post, I’ll complete the proof by handling Case 4. Before starting Case 4, I like to take inventory of where we stand in the proof at this point. We have now proven the theorem for all positive rational numbers and for all negative rational numbers. There’s only one rational number left: $x = 0$. And this single case is simply handled through Property 1: $f(a^0) = f(1) = 0$

I also like to keep track of which hypotheses have been used so far in the proof. A quick review of Cases 1-3 will reveal that Properties 1-3 have all been used at least once, but Property 4 (the assumption that $f$ is continuous) has not be used so far. Therefore, we had better expect to use it before completing the proof.

I won’t tell the class this (for fear of discouraging them), but the proof of Case 4 is a bit more abstract than Cases 1-3. I can give a numerical example that (hopefully) will shed some insight into the actual proof. However, for Case 4, the actual proof will not be a perfect parallel of the numerical example (as in Cases 1-3).

Idea behind Case 4. Let’s pick a familiar irrational number like $\sqrt{2}$. There is a natural way to approximate $\sqrt{2}$ by a sequence of rational numbers… namely, the sequence of numbers obtained by taking one extra digit in the decimal expansion of $\sqrt{2}$. In other words, $r_1 = 1$ $r_2 = 1.4$ $r_3 = 1.41$ $r_4 = 1.414$

and so on.

In this way, $\displaystyle \lim_{n \to \infty} r_n = \sqrt{2}$.

We would hope that the sequence $f \left( a^1 \right), f \left( a^{1.4} \right), f \left( a^{1.41} \right), f \left( a^{1.414} \right), \dots$

converges to the obvious limit of $f \left( a^{\sqrt{2}} \right)$.

However, this sequence is also equal to $1, 1.4, 1.41, 1.414, \dots$

since each exponent is rational. Since a sequence has only one limit, we conclude that these two limits should be equal: $\sqrt{2} = f \left( a^{\sqrt{2}} \right)$

So that’s the idea of the formal proof, which we now tackle. In the proof below, I’ve marked with quotations some of the more parenthetical steps so that the main argument of the proof stands out a little bit better. You’ll notice that, unlike Cases 1-3, I don’t use as much directed questioning to get students to volunteer the next step of the proof with minimal assistance from me. That’s because I haven’t figured out a good way to use inquiry to quickly get through Case 4.

Proof of Case 4. Let $\{ r_n \}$ be a sequence of rational numbers that converges to $x$. (Parenthetically, I’ll mention that the sequence of decimal approximations would be one such sequence, just to make this mysterious $\{ r_n \}$ thing that just appeared out of the blue a little less daunting. Of course, any sequence of rational numbers that converges to $x$ will do. Therefore, $f \left( a^x \right) = f \left( a^{\lim_{n \to \infty} r_n} \right)$

The function $g(x) = a^x$ is continuous. From the ordinary definition of continuous used in calculus, this means that $\displaystyle \lim_{x \to c} g(x) = g(c)$.

In other words, the function and the limit can be interchanged. (I’ll usually throw in my standard joke about functions commuting at this point in the lecture.) Stated in terms of a sequence $r_n \to x$, this means that $\displaystyle \lim_{n \to \infty} g(r_n) = g(x) = g \left( \lim_{n \to \infty} r_n \right)$.

Stated another way, $\displaystyle \lim_{n \to \infty} a^{r_n} = a^{ \lim_{n \to \infty} r_n}$.

In light of the above work, we conclude that $f \left( a^x \right) = f \left( a^{\lim_{n \to \infty} r_n} \right) = f \left( \displaystyle \lim_{n \to \infty} a^{r_n} \right)$

Stated simply, the function and the limit interchange.

We now perform a similar step for the function $f$. Because $f$ is assumed to be continuous, we know that $\displaystyle \lim_{n \to \infty} f(s_n) = f(c) = f \left( \lim_{n \to \infty} s_n \right)$

if $\{ s_n \}$ is a sequence that converges to $c$. So, if we replace $s_n$ by $a^{r_n}$ and $c$ by $\displaystyle \lim_{n \to \infty} a^{r_n}$, we conclude that $\displaystyle \lim_{n \to \infty} f \left( a^{r_n} \right) = f \left( \displaystyle \lim_{n \to \infty} a^{r_n} \right)$

From the above insight, we see that we have the next step of the proof: $f \left( a^x \right) = f \left( a^{\lim_{n \to \infty} r_n} \right)$ $= f \left( \displaystyle \lim_{n \to \infty} a^{r_n} \right)$ $= \displaystyle \lim_{n \to \infty} f \left(a^{r_n} \right)$

From now on, the concluding steps are pretty straightforward. The exponent on the last line is a rational number. Therefore, by Cases 2 and 3, we have produce the next step: $f \left( a^x \right) = f \left( a^{\lim_{n \to \infty} r_n} \right)$ $= f \left( \displaystyle \lim_{n \to \infty} a^{r_n} \right)$ $= \displaystyle \lim_{n \to \infty} f \left(a^{r_n} \right)$ $= \displaystyle \lim_{n \to \infty} r_n$

Finally, by definition from the top of the proof, we can evaluate this limit: $f \left( a^x \right) = f \left( a^{\lim_{n \to \infty} r_n} \right)$ $= f \left( \displaystyle \lim_{n \to \infty} a^{r_n} \right)$ $= \displaystyle \lim_{n \to \infty} f \left(a^{r_n} \right)$ $= \displaystyle \lim_{n \to \infty} r_n$ $= x$

This concludes the proof that $f \left( a^x \right) = x$, even if $x$ is an arbitrary (possibly irrational) real number.