Did my friend remember this exponent question wrong?

Factual Questions
1

I promise this isn’t homework, I’m 30 years old.

A friend of mine just took a test where she encountered this problem, as she recalls it:

20[sup]5[/sup] = 2[sup]m[/sup] + 2[sup]n[/sup], solve for mn

without using logs. Although even with logs, I’m not sure how you would do it, because of the +. And then even if she misremembered a + where it was actually a *, I’m still not sure, because I can’t seem to match up the bases, which is how you normally do this (typically simple) type of question.

Any ideas?

2

Was it a product rather than a sum?
n=2
m=3
Seems to be the obvious setup.

3

That problem doesn’t have a solution for any integer pair (m, n). If you have a number whose binary representation matches 1010 (i.e., a 1, zero or more instances of 0, another 1, and zero or more instances of 0), then you’d solve it by figuring out what those two powers of two are. For instance, 68[sub]10[/sub] = 1000100[sub]2[/sub] = 1000000[sub]2[/sub] + 100[sub]2[/sub] = 2[sup]6[/sup] + 2[sup]2[/sup].

4

It can’t be a product, because 20 is divisible by 5 and 2[sup]x[/sup] is not for any x.

5

:smack: Screwed that up didn’t I. My apologies.

6

Further, if you allow m and n to be real numbers, you can solve for m in terms of n (yes, using logs), and it’s pretty obvious that the product mn will not be the same for two such solutions. So yeah, the problem is ill-posed as stated.

7

Ok, thanks for the replies. She probably just remembered something wrong.

8

I recall a similar problem from my first math competition. (It was a one-point "warmup’ question): if 2[sup]5[/sup] = 2[sup]m[/sup] * 2[sup]n[/sup], what is m+n?

The sum/product are switched vs your example, and the first term has a 2, not a 20,
but m+n=5 for all real m, n – pretty much by the definition of ‘powers’.

This problem could be stated with 20[sup]5[/sup], as well. 20[sup]5[/sup] is a constant.
and for any C= 2[sup]m[/sup]*2[sup]n[/sup], m+n=log[sub]2/sub

In fact, were I judging, I’d accept an algebraic answer as well as a numerical approximation
(unless otherwise specified) because it showed a grasp of the principle.
log[sub]2/sub
5 log[sub]2/sub
~21.61
etc.