## Interesting Exponent Problem

*February 17, 2013 at 5:38 pm* *
2 comments *

In the expression 3*x*^{2}, the 2 is called an *exponent*. And in the expression 4*y*^{3}, the 3 is called the *y*‑ponent.

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

If 2

^{m}= 8 and 3^{n}= 9, what is the value of m · n?

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

If 2

^{m}= 9 and 3^{n}= 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.

[…]

Ah, good. Glad you’re back.

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 2^{m} = 64, then *m* = 6. This would mean that 3^{n} = 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 3^{1} ≠ 9/8.

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

If 2

^{m}= k and 3^{n}= 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 2

^{m}= p and 3^{n}= q?

If 2

^{m}= p and 3^{n}= 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 2^{m} = 9, then 2^{mn} = 9^{n} = 3^{2n} = 8^{2} = 2^{6}, so *mn* = 6.

My colleague Al Goetz solved it as follows.

Since

Then

Entry filed under: Uncategorized. Tags: exponents, logarithm.

1.Larry Davidson | February 17, 2013 at 6:07 pmYou 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 pmD’oh! Thanks, Larry. I corrected it in the post.