An exponential model (solving with ln)

Exponentials and logarithms

A-level Maths (9MA0)

ORIGINAL

Use the starting value for A, take ln of both sides to find k, then keep k exact when you solve for the time.

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Question

The number of bacteria,

N

, in a culture

t

hours after it is set up is modelled by

N = Ae^{kt}

, where

A

and

k

are constants. Initially there are 500 bacteria, and after 3 hours there are 1350. (a) Find the value of

A

and show that

k = \tfrac{1}{3}\ln 2.7

. (b) Find, to 3 significant figures, the time in hours at which the model predicts there will be 5000 bacteria.

Every step worked, with the reasoning.

1
$t = 0$ : $500 = Ae^{0} = A$ , so $A = 500$
$e^0 = 1$ , so the initial value is $A$ .
2
$t = 3$ : $1350 = 500e^{3k}$ , so $e^{3k} = 2.7$
Substitute the second data point and divide by 500.
3
$3k = \ln 2.7$ , so $k = \tfrac{1}{3}\ln 2.7 = 0.331$ (3 s.f.)
Take natural logs of both sides: $\ln$ undoes $e$ .
4
$5000 = 500e^{kt}$ , so $e^{kt} = 10$
Set $N = 5000$ and divide by 500.
5
$kt = \ln 10$ , so $t = \dfrac{\ln 10}{k} = \dfrac{3\ln 10}{\ln 2.7} = 6.954\ldots$
Take logs again and use the exact $k$ : rounding $k$ to 0.331 first gives $6.956\ldots$ , which rounds to the wrong answer, 6.96.

Answer: (a) $A = 500$ , $k = \tfrac{1}{3}\ln 2.7 \approx 0.331$ (b) $t = 6.95$ hours (3 s.f.)