Interesting Exponent Problem

In the expression 3x2, the 2 is called an exponent. And in the expression 4y3, the 3 is called the y‑ponent.

Speaking of exponents, here’s a problem you’ll have no trouble solving.

If 2m = 8 and 3n = 9, what is the value of m · n?

This next one looks similar, but it’s a bit more difficult.

If 2m = 9 and 3n = 8, what is the value of m · n?

One way to solve this problem is to compute the values of m and n using a logarithm calculator, though there’s an algebraic approach that’s far more elegant. (I’ll provide the algebraic solution at the end of this post, so as not to spoil your fun.)

Take a few minutes to solve that problem. Go ahead, give it a try. I’ll wait for you.

[…]

Are you surprised by the result? I sure was.

It made me wonder, “If the product of the two numbers is 72, will it always be the case that m · n = 6?”

It only took me a minute to realize that the answer is no, which can be seen with numerous counterexamples. For instance, if 2m = 64, then m = 6. This would mean that 3n = 9/8, because 64 × 9/8 = 72, but it would also mean that n = 1, since m · n = 6. That is obviously a contradiction, since 31 ≠ 9/8.

If the product of the two numbers is held constant at 72, then the problem looks something like this:

If 2m = k and 3n = 72/k, what is the value of m · n?

The following graph shows the result for 1 < k < 72.

So that leads to a couple of follow-up questions.

For what integer values of p, q will p · q = 72 and m · n = 6, if 2m = p and 3n = q?

If 2m = p and 3n = q, for what integer values of p, q will m · n = 6 when m, n are not integers?

The answers, of course, are left as an exercise for the reader.

Here’s the algebraic solution to the problem above.

Since 2m = 9, then 2mn = 9n = 32n = 82 = 26, so mn = 6.

My colleague Al Goetz solved it as follows.

Since

$m \log{2} = \log{9}$

$n \log{3} = \log{8}$

Then

$m \cdot n = \frac{\log{9}}{\log{2}} \cdot \frac{\log{8}}{\log{3}} = = \frac{\log{3^2} \cdot \log{2^3}}{\log{2} \cdot \log{3}} = \frac{2 \log{3} \cdot 3 \log{2}}{\log{2} \cdot \log{3}} = 6$

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• 1. Larry Davidson  |  February 17, 2013 at 6:07 pm

You have a typo: you copied and pasted the original problem, so you have the same problem twice. You forgot to swap the exponents!

• 2. venneblock  |  February 17, 2013 at 8:27 pm

D’oh! Thanks, Larry. I corrected it in the post.

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