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A certain species of bacteria grows at a constant rate, its [#permalink]
22 Oct 2013, 13:21

A certain species of bacteria grows at a constant rate, its numbers increasing by an unknown multiplier every hour. At the end of one hour (time t = 1), the bacteria colony’s population size is p; at the end of five hours (t = 5), the colony’s population has grown to p^2.

In the first column, indicate the factor by which the number of bacteria grows each hour. In the second column, indicate the population at the start of the time period described (at time t = 0). Make only two selections, one in each column.

First Column A) p^1/4 B) p^1/3 C) p^1/2 D) p^3/4 E) p F) p^3/2

Second column

A) p^1/4 B) p^1/3 C) p^1/2 D) p^3/4 E) p F) p^3/2

I am ADAMANT that the OA/OE on this question is incorrect

This short problem is quite difficult because of the subject matter. The first sentence describes a value that is increasing steadily “by an unknown multiplier” or growth factor. So there is some multiplier that could be used to calculate the size of the population at the end of every hour. For example, if you know that a population starts out with 2 bacteria and the multiplier is 5, then at the end of the first hour, there will be 2 × 5 = 10 bacteria; at the end of the second hour, there will be 10 × 5 = 50 bacteria, and so on.

The second sentence then provides information to calculate that unknown multiplier, or growth factor; let’s call that multiplier r. At t = 1, the population is p. At t = 5, the population is p2. You are told that the growth factor corresponds to hourly growth, so there are 4 one-hour intervals in the timeframe described. So the question becomes this: what value of r, multiplied by itself 4 times (for the 4 one-hour intervals), would take you from p to p2?

Multiplying r by itself 4 times is the same thing as raising r to the 4th power: (r)(r)(r)(r) = r4. In the four hours described, the bacteria grew from p to p2 (or p times p) so the bacteria grew by a factor of p. Therefore, p = r4, or p1/4 = r. (“p to the one-fourth power” is the same thing as the fourth root of p.)

The hourly growth factor, r, is p1/4, or 4√p.

You can now use the growth factor, p1/4, to calculate the population size at t = 0. If you were going from hour 1 to hour 2, you would multiply the hour 1 population by the growth factor in order to get to the hour 2 population (as in the above table). Because you are working backwards, from hour 1 to hour 0, you instead divide hour 1’s population by the growth factor:

p / p1/4 = p1 – 1/4 = p3/4

“p to the 3/4 power” is the fourth root of p3, or 4√(p3).

Column 1: The correct answer is A.

Column 2: The correct answer is D.

my bone to pick comes from the fact that they're only multiplying that base number, not accounting for exponential growth

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)