Question

answer each question and submit your work in the space provided.
the number of bacteria present in a culture after t minutes is given as b = 100e^{kt}. if there are 4331 bacteria present after 4 minutes, find k. (10 points)

Explanation:

Step1: Substitute the given values into the formula

We are given the formula \(B = 100e^{kt}\), with \(B = 4331\) and \(t = 4\). Substituting these values gives \(4331=100e^{4k}\).

Step2: Isolate the exponential term

Divide both sides of the equation by \(100\): \(\frac{4331}{100}=e^{4k}\), so \(43.31 = e^{4k}\).

Step3: Take the natural logarithm of both sides

Using the property \(\ln(e^{x})=x\), we take \(\ln\) of both sides: \(\ln(43.31)=\ln(e^{4k})\). Since \(\ln(e^{4k}) = 4k\), the equation becomes \(\ln(43.31)=4k\).

Step4: Solve for \(k\)

We know that \(\ln(43.31)\approx3.761\). Then \(k=\frac{\ln(43.31)}{4}\). Substituting the value of \(\ln(43.31)\), we get \(k=\frac{3.761}{4}=0.94025\approx0.94\).

Answer:

\(k\approx0.94\)