A researcher is studying the exponential growth of a bacterial culture. The growth is modelled by the equation \(y = Ae^{kx}\), where \(y\) represents the population at time \(x\), \(A\) is the initial population and \(k\) is the growth rate.
(a) Apply the natural logarithm to both sides of the equation to show that the model can be linearized into the form \(\ln(y) = \ln(A) + kx\).
(b) Using the data points \((x, y) = (0, 150)\) and \((2, 450)\), determine the slope of the line obtained after the transformation. Interpret this slope in the context of exponential growth.
Question
Given the exponential model \(y = Ae^{kx}\), apply the natural logarithm to both sides of the equation to derive a linear equation in the form \(\ln(y) = \ln(A) + kx\).
Hint
Hint
Hint
Remember that \(\ln(e^{kx}) = kx\).
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Answer
Answer
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Explanation
Explanation
- Formula Booklet
Complete Answer
Taking the natural logarithm of both sides yields \(\ln(y) = \ln(A) + kx\).
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Answer Explanation
Answer Explanation
Question
Using the linearized equation \(\ln(y) = \ln(A) + kx\) and the data points \((x, y) = (0, 150)\) and \((2, 450)\), calculate the slope of the line, which represents the growth rate \(k\). Interpret this slope in the context of exponential growth.
Hint
Hint
Hint
Subtract \(\ln(150)\) from \(\ln(450)\) and divide by the change in \(x\) (which is 2).
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Answer
Answer
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Explanation
Explanation
- Formula Booklet
Complete Answer
The slope is \(k = \frac{\ln(450) – \ln(150)}{2} = \frac{\ln(3)}{2} \approx 0.549\). The growth rate is approximately 0.549.
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Answer Explanation
Answer Explanation
