Interesting Exponent Problem

February 17, 2013 at 5:38 pm 2 comments

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.

[…]

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 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.

2^m Graph

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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Blog: Proofs from the Book The Math of Maker’s Mark

2 Comments Add your own

  • 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!

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

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

      Reply

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The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

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