Question:

The bacteria population in a laboratory culture is modeled by the exponential growth equation $P(t) = P_0 \cdot e^{kt}$, where $P(t)$ is the population at time $t$ (measured in hours), $P_0$ is the initial population, and $k$ is a constant.

Given that a bacterial culture initially has 1000 bacteria and that the population triples after 2 hours, find:

a) The value of $k$, representing the exponential growth rate. b) An expression for $P(t)$, the population at any time $t$ in hours. c) The population after 6 hours.

Answer:

a) To find the value of $k$, we can use the given information that the population triples after 2 hours.

We know that $P(2) = 3P_0$, where $P(2)$ represents the population at 2 hours and $3P_0$ is three times the initial population.

Substituting the given values, we have:

$3P_0 = P_0 \cdot e^{2k}$

Dividing both sides by $P_0$:

$3 = e^{2k}$

Taking the natural logarithm (ln) of both sides:

$\ln(3) = \ln(e^{2k})$

Using the property $\ln(a^b) = b \ln(a)$:

$\ln(3) = 2k \ln(e)$

Since $\ln(e) = 1$, we have:

$\ln(3) = 2k$

Now, solving for $k$:

$k = \frac{\ln(3)}{2}$

Therefore, the value of $k$ is equal to $\frac{\ln(3)}{2}$.

b) With the value of $k$ determined, we can now find an expression for $P(t)$, the population at any time $t$.

Using the equation $P(t) = P_0 \cdot e^{kt}$, we substitute our value of $k$ into the equation:

$P(t) = P_0 \cdot e^{\frac{\ln(3)}{2} \cdot t}$

Simplifying further, we can rewrite it as:

$P(t) = P_0 \cdot (e^{\ln(3)})^{t/2}$

Using the property $e^{\ln(a)} = a$:

$P(t) = P_0 \cdot 3^{t/2}$

Therefore, an expression for $P(t)$, the population at any time $t$, is $P(t) = P_0 \cdot 3^{t/2}$.

c) To find the population after 6 hours, we substitute $t = 6$ into the expression for $P(t)$.

$P(6) = P_0 \cdot 3^{6/2}$

$P(6) = P_0 \cdot 3^3$

$P(6) = P_0 \cdot 27$

Since the initial population, $P_0$, is given as 1000 bacteria, we have:

$P(6) = 1000 \cdot 27$

$P(6) = 27000$

Therefore, the population after 6 hours is 27,000 bacteria.