To find the expression for population growth, we'll use the given exponential model formula, P(t) = (P0)(e^kt), where P(t) represents the population at time t, P0 represents the initial population, e is the mathematical constant approximately equal to 2.71828, and k represents the growth rate.
In this scenario, the initial population (P0) is 500, and the time for the population to double (t) is 15 minutes. To find the growth rate (k), we can use the fact that the population doubles every 15 minutes.
First, let's substitute the known values into the formula: P(t) = (500)(e^kt). Now we need to solve for k.
Since the population doubles every 15 minutes, we can write the equation: 2 = e^(k * 15).
To solve for k, we need to take the natural logarithm (ln) of both sides of the equation to eliminate the exponential term:
ln(2) = ln(e^(k * 15)).
Using the property of logarithms (ln(a^b) = b * ln(a)), we can rewrite the equation as:
ln(2) = k * 15 * ln(e).
Since ln(e) is equal to 1, the equation simplifies to:
ln(2) = k * 15.
Now, we can solve for k by dividing ln(2) by 15:
k = ln(2) / 15.
The value of k is approximately 0.0462.
Finally, we can substitute the values of P0, t, and k into the formula to find the population at 87 minutes:
P(t) = (P0)(e^kt) = 500 * (e^(0.0462 * 87)) ≈ 500 * e^(4.0146) ≈ 500 * 55.6184 ≈ 27,809.
Therefore, the population of E. coli bacteria would be approximately 27,809 after 87 minutes.