A certain species of bacteria grows at a constant rate, its numbers in

Dear AccipiterQ,

I'm happy to help. I believe the OA given is correct, and does correctly account for exponential growth. First, here's a blog on exponents:
https://magoosh.com/gmat/2012/exponent-p ... -the-gmat/

Part of what's tricky is that the start time, with population p, is at t = 1, and the the later time, at t = 5, is FOUR hours later. That's the first tricky thing.

Also, keep in mind, for exponential growth, of course the size of the population doesn't increase linearly, but the exponent on the function does increase linearly. Thus, at time #1, the population is a P^A, and at time #2, it's at P^B (where B>A), then at a time exactly between those two times, the population will have an exponent exactly between those two values (a.k.a. the average), P^[(A+B)/2]

Part one: Let r be the unknown hourly growth ratio:
t = 1 --> p
t = 2 --> p*r
t = 3 --> p*r*r = p*(r^2)
t = 4 --> p*r*r*r = p*(r^3)
t = 5 --> p*r*r*r*r = p*(r^4)
Set that last value equal to p^2
p*(r^4) = p^2
r^4 = p
r = p^(1/4)
Column #1, choice = (A)

Given that
t = 1 --> p
t = 0 --> p/r = p/[p^(1/4)] = p^(3/4)
Column #2, choice = (D)

Does all this make sense?
Mike