Different definitions of logarithm (Part 4)

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 2 during class time. Cases 3-4 will appear in the coming posts.

green line

Idea behind Case 2. 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

2 = f(a^2)

I’ll ask, “How do we know this is true?” The immediate answer: We just did Case 1. I’ll then do something a little unusual and rewrite this equation in a more complicated way:

2 = f(a^2) = f \left( \left[a^{2/3} \right]^3 \right)

After double-checking that the class agrees with this step (even if I just made the right-hand more complicated instead of the usual step of simplifying the right-hand side), I’ll then ask, “OK, we have something to the third power. What can we now do to the right-hand side?” Almost immediately, someone will volunteer the correct next steps using Property 3:

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

I’ll then ask, “How can we simplify the right-hand side?” After a moment of thought, someone will volunteer the correct next step:

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

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

 I’ll then ask, “How do we isolate the f \left( a^{2/3} \right) term?” The obvious correct answer:

\displaystyle \frac{2}{3} = f(a^{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}.

The natural next question is, “Can we do this for any positive rational number and not just \frac{2}{3}?” This leads to the proof of Case 2. 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 2. Let x = \displaystyle \frac{m}{n} where m, n \in \mathbb{R}^+. Then

m = f(a^m)

m = f \left( \left[ a^{m/n} \right]^n \right)

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

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

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

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

Of course, the special case x = \frac{2}{3} is not logically necessary to prove Case 2. Though not logically necessary, I’ve found it to be pedagogically convenient. From the school of hard knocks, I’ve found that the proof of Case 2 goes over easier with students when they see the idea of the proof presented concretely and then abstractly.

2 thoughts on “Different definitions of logarithm (Part 4)

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.