Question

problem 1. a bacteria population grows according to the function $f(t)=100e^{kt}$, where $t$ is in hours. after 3 hours, the bacteria population is 400. answer the following questions:
(a) what is the initial size of the bacteria population.
(b) find the growth rate of the bacteria (ie: find $k$ in the function above).
(c) what does the model predict the size of the population would be after 5 hours?

Explanation:

Step1: Find the initial size

For the function \(f(t)=100e^{kt}\), when \(t = 0\) (initial time), \(f(0)=100e^{k\times0}\).
Since \(e^{0}=1\), then \(f(0)=100\).

Step2: Find the growth rate \(k\)

We know that when \(t = 3\), \(f(3)=400\). Substitute into the function: \(400=100e^{3k}\).
Divide both sides by 100: \(4 = e^{3k}\).
Take the natural logarithm of both sides: \(\ln(4)=\ln(e^{3k})\).
Using the property \(\ln(e^{x})=x\), we get \(\ln(4) = 3k\).
Since \(\ln(4)=2\ln(2)\), then \(k=\frac{2\ln(2)}{3}\approx\frac{2\times0.693}{3}= 0.462\).

Step3: Predict the size after 5 hours

Substitute \(t = 5\) and \(k=\frac{2\ln(2)}{3}\) into \(f(t)=100e^{kt}\).
\(f(5)=100e^{\frac{2\ln(2)}{3}\times5}=100e^{\frac{10\ln(2)}{3}}\).
Using the property \(a\ln(b)=\ln(b^{a})\) and \(e^{\ln(x)}=x\), we have \(f(5)=100\times2^{\frac{10}{3}}\).
\(2^{\frac{10}{3}}=\sqrt[3]{2^{10}}=\sqrt[3]{1024}\approx10.08\).
So \(f(5)=100\times10.08 = 1008\).

Answer:

(a) The initial size of the bacteria population is \(100\).
(b) The growth rate \(k\) is \(\frac{2\ln(2)}{3}\approx0.462\).
(c) The size of the population after \(5\) hours is approximately \(1008\).