# Different definitions of logarithm (Part 3)

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 describe how I prompt my students to prove Case 1 during class time. Cases 2-4 will appear in the coming posts. Idea behind Case 1. Though not formally necessary for the proof, I’ve found it helpful to illustrate the idea of the proof with a specific example before proceeding to the general case. So — on the far end of the chalkboard, away from the space that I’ve allocated for the formal write-up of the proof — I’ll write $f(a^4) =$

I’ll then ask, “How else can we write $a^4$?” Someone will usually suggest $a \cdot a \cdot a \cdot a$, and so I’ll write this as the next step: $f(a^4) = f(a \cdot a \cdot a \cdot a)$

I’ll then ask, “OK, we have a product here. How can we simplify the right-hand side?” After a moment of thought, someone will volunteer that Property 3 allows the right-hand side to be split up into pieces: $f(a^4) = f(a \cdot a \cdot a \cdot a) = f(a) + f(a) + f(a) + f(a)$

(Technically, this requires mathematical induction to generalize Property 3 from a product of two numbers to a product of arbitrarily many numbers, but I don’t think that it’s worth the time to expound on this pedantic point.) I’ll then ask, “How can we simplify this?” Almost immediately, someone will usually volunteer Property 2: $f(a^4) = f(a \cdot a \cdot a \cdot a) = f(a) + f(a) + f(a) + f(a) = 1 + 1 + 1 + 1 = 4$

I’ll then note that we’ve finished what we set out to do: show that $f(a^x) = x$ when $x = 4$.

The natural next question is, “Can we do this for any positive integer and not just 4?” This leads to the proof of Case 1. I’ve found that it’s helpful to walk through this proof line by line in step with the case of $x=4$, so that students can see how the steps of this more abstract proof correspond to the concrete example of $x =4$.

Proof of Case 1. $f(a^m) = f(a \cdot a \cdot \dots \cdot a)$ $= f(a) + f(a) + \dots + f(a)$ $= 1 + 1 + \dots + 1$ $= m$

Of course, the special case $x = 4$ is not logically necessary to prove Case 1. That said, from the school of hard knocks, I’ve found that the proof of Case 1 goes over easier with students when they see the idea of the proof presented concretely and then abstractly.

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