GPrep 1 Question

[Nov1907]

1. For how many integers k is k^2 = 2^k ?
(A) None
(B) One
(C) Two
(D) Three
(E) More than three

I was able to figure this out by trial and error - (works for k = 2 and 4 and k^2 is never large enough to reach 2^k for any other powers of 2.). Is there a more mathematical message to do this without using the log function? Would appreciaete the help.

[RonPurewal]

Wow, really? They asked you that?

A more systematic approach: First, rule out negative numbers and 0.
* 2^(negative integer) is between 0 and 1, whereas (negative integer)^2 is greater than 1. So, no negatives work.
* 2^0 = 1, 0^2 = 0, so 0 doesn't work either.

Positive numbers:
* Try 1, 2, 3, 4, 5, and notice that 2 and 4 happen to work.
- Notice that 2^5 is bigger than 5^2
- Notice that 2^(next consecutive integer) is always double the number you just found, whereas the next perfect square is never even close to double the last one
- So, 2^k will grow at a rate that outpaces that of k^2, and never again the twain shall meet

I gather that this is pretty much what you did, but with a little more formality thrown in.

[Nov1907]

Thanks Ron. I pretty much neglected the -ve numbers and zero (need to stop doing that quickly, really hurts on some of those number substitution inequalities). Guess I lucked out on this question.

[RonPurewal]