Different definitions of logarithm (Part 5)

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 3 during class time. Cases 4 will appear in tomorrow’s post.

green line

Idea behind Case 3. 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 \left( a^{-2/3} \cdot a^{2/3} \right) =

I’ll then ask, “How else can we simplify the left-hand side?” As we’ll see below, there are actually two legitimate ways of proceeding. Someone will usually suggest just simplifying the product, and so I’ll write this as the next step:

f \left( a^{-2/3} \cdot a^{2/3} \right) = f \left( a^0 \right)

I’ll then ask a very open-ended question, “Now what?” Usually, someone will suggest simplifying the right-hand side using Property 1:

f \left( a^{-2/3} \cdot a^{2/3} \right) = f \left( a^0 \right) = 0

By this point, after completing Cases 1 and 2, someone will usually suggest expanding the left-hand side:

f \left( a^{-2/3} \right) + f \left( a^{2/3} \right) = 0

I’ll then ask, “What can we do now?” Hopefully, someone will observe that the second term can be simplified using Case 2:

f \left( a^{-2/3} \right) + \displaystyle \frac{2}{3} = 0

f \left( a^{-2/3} \right) = - \displaystyle \frac{2}{3}

I’ll then note that we’ve finished what we set out to do: show that f(a^x) = x when x = - \frac{2}{3}, a negative rational number.

The natural next question is, “Can we do this for any negative rational number and not just -\frac{2}{3}?” This leads to the proof of Case 3. I’ve found that it’s helpful to walk through this proof line by line in step with the case of x= -\frac{2}{3}, so that students can see how the steps of this more abstract proof correspond to the concrete example of x = =\frac{2}{3}.

Proof of Case 3. Let m, n \in \mathbb{Z}^+. Then

f \left(a^{-m/n} \cdot a^{m/n} \right) = f \left(a^0 \right)

f \left( a^{-m/n} \right) + f \left( a^{m/n} \right) = 0

f \left( a^{-m/n} \right) + \displaystyle \frac{m}{n} = 0

f \left( a^{-m/n} \right) = - \displaystyle \frac{m}{n}

Again, I’ve found that the special case x = - \frac{2}{3} is pedagogically helpful, if not logically necessary to prove Case 3.

2 thoughts on “Different definitions of logarithm (Part 5)

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