The population of a colony of bacteria doubles every 15 minutes. The population of the col… — SAT

The population of a colony of bacteria doubles every $15$ minutes. The population of the colony, $P(t)$ , after $t$ hours can be modeled by the function $P(t) = P_0 \cdot 8^{kt}$ , where $P_0$ is the initial population of the colony and $k$ is a constant. What is the value of $k$ ?

A
$\frac{1}{12}$
B
$\frac{3}{4}$
$\frac{4}{3}$ Answer
D
$12$

Answer

The correct value of

k

\frac{4}{3}

Since the population doubles every

15

minutes, it doubles

4

times per hour. Therefore, the population after

t

hours is

P_0 \cdot 2^{4t}

. To express this model in the form

P_0 \cdot 8^{kt}

, we equate the growth factors:

2^{4t} = 8^{kt}

. Since

8 = 2^3

, this can be rewritten as

2^{4t} = (2^3)^{kt} = 2^{3kt}

. Setting the exponents equal gives

4t = 3kt

, which simplifies to

4 = 3k

and yields

k = \frac{4}{3}

Step-by-Step Solution

Determine the number of doubling periods per hour.

The population doubles

4

times per hour.

Since there are

60

minutes in an hour and the population doubles every

15

minutes, the number of doubling periods in one hour is

\frac{60}{15} = 4

Formulate the population growth equation in terms of base 2.

P(t) = P_0 \cdot 2^{4t}

If the population doubles

4

times per hour, then after

t

hours, it has doubled

4t

times, representing a growth factor of

2^{4t}

Equate the growth model to the given function and express using a common base.

2^{4t} = (2^3)^{kt} \implies 2^{4t} = 2^{3kt}

Set the two expressions for the population equal:

P_0 \cdot 2^{4t} = P_0 \cdot 8^{kt}

. Dividing by

P_0

gives

2^{4t} = 8^{kt}

. Since

8 = 2^3

, rewrite the right side to have a base of

2

Solve for the constant

k

by equating the exponents.

4t = 3kt \implies 4 = 3k \implies k = \frac{4}{3}

Since the bases on both sides of the equation are equal, their exponents must also be equal. Divide both sides by

t

(since

t > 0

) and isolate

k

Key Concept

Converting bases in exponential functions and adjusting exponents for different units of time.

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