Question

A biologist is researching a newly discovered species of bacteria. At

time t = 0, he puts 200 bacteria into what he has determined to be a favorable growth. Eight hours later, he measures 800 bacteria. This can
be modeled by the equation y = aekt, where a is the initial amount of
bacteria. Find the rate of growth, k, of the bacteria.

Answer 1

k ≈ 0.1733

Explanation:

Using the numbers given, we find the multiplier of the population is 800/200 = 4 in 8 hours. This means the equation of growth can be written ...

y = 200(4^(t/8))

When this is written in the form ...

y = 200·e^(kt)

We can compare the two equations to see that ...

4^(1/8) = e^k

Taking natural logarithms, we have ...

1/8·ln(4) = k ≈ 0.1733

The growth rate k is approximately 0.1733.

_____

Additional comment

The comparison we were looking for was (4^(1/8))^t = (e^k)^t. Instead of "comparing" the equations, you could set them equal and solve for k.

200(4^(t/8)) = 200e^(kt)

4^(t/8) = e^(kt) . . . divide by 200

t/8·ln(4) = kt·ln(e) . . . . take natural logs

1/8·ln(4) = k . . . . . . use ln(e) = 1 and divide by t