Problem involving geometric progression

Question

Question:
The bacteria in a certain culture double every $7.3$ hours. The culture has $7,500$ bacteria at the start. How many bacteria will the culture contain after $3$ hours?

Possible Answers:
a. $9,449$ bacteria
b. $9,972$ bacteria
c. $40,510$ bacteria
d. $8,247$ bacteria


I got this answer right on my quiz but I want to be sure that I can do it again. Please help me with setting up this problem and getting the correct answer.

Answer 1

If you have any question about this formula, you may think of the following:

Let $x_0$ be the initial population. We know that every $7.3$ hours the population doubles itself. Taking advantage of that, we have that: $$\begin{array}{ccl} \text{ After $(1 \times 7.3)$ hours the population becomes} & \to & 2 x_0\\ \text{ After $(2\times 7.3)$ hours the population becomes } & \to & 2\cdot (2x_0) = 2^2 \cdot x_0\\ \text{ After $(3\times 7.3)$ hours the population becomes} & \to & 2 \cdot ( 2^2\cdot x_0) = 2^3\cdot x_0\\ \vdots\\ \text{ After $(n\times 7.3)$ hours the population becomes} &\to & 2^n \cdot x_0 \end{array}$$ So, we want to find the population after $3$ hours, which means $n = \frac{3}{7.3}$.

Thus, the population after $3$ hours will be $2^{3/7.3}\cdot 7500$.

Answer 2

The rate of growth for the bacteria can be described by the following formula:

$$n_{t} = 7500 \cdot 2^{t/7.3}$$

...where $n_{t}$ is the number of bacteria at time $t$.

Let $t=3$ and you will find your answer.

Derivation of the formula:

Let $n_i$ represent the initial number of bacteria.

Let $b$ represent the base of the exponential function, which is the rate at which the bacteria reproduce. For example, if the bacteria double every $t$ hours then $b=2$. If the bacteria triple every $t$ hours then $b=3$, etc.

We have an exponential equation of the form:

$$n_t = n_i \cdot b^a$$

Exponentiation of $b$ must change with regard to the amount of time passing (represented by $t$). In this case, we are told that the bacteria doubles after 7.3 hours, so let $a = 1$ at $t=7.3$

$$n_{7.3} = 7500 \cdot 2^1 = 7500 \cdot 2^{7.3/7.3}$$

Clearly, the numerator in the exponent is the deciding factor of how many bacteria are predicted to be present at any given time. We would expect less bacteria to be present after 3 hours than after 7.3 hours. The formula $n_{t} = 7500 \cdot 2^{t/7.3}$ describes this rate-of-growth.