Question

In a research lab, a cell biologist is growing a strain of bacteria under two different conditions. Culture A had an initial population of 200 and doubles every hour. Culture B had an initial population of 819200 but has been contaminated. Its population is now decreasing by half every hour. When will the two cultures have an equivalent population? Solve using an exponential equation.

Answer 1

They will have the same population after 6 hours

Explanation:

The general exponential equation format for this is;

P = P_o(e^(kt))

For Culture A, we are told that it had an initial population of 200 and doubles every hour.

Thus;

After 1 hour, P = 400

After 2 hours, P = 800

After 3 hours, P = 1600

Thus;

At t = 1, we have;

400 = 200(e^(k × 1))

400/200 = e^(k)

e^(k) = 2 - - - (eq 1)

At t = 2, we have;

800 = 200(e^(k × 2))

800/200 = e^(2k)

e^(2k) = 4 - - - (eq 2)

To find k, let's divide eq 1 by eq 2.

(e^(2k))/e^(k) = 4/2

e^(2k - k) = 2

e^(k) = 2

k = In 2

k = 0.6931

Thus;

P = 200e^(0.6931t)

For Culture B, we are told that it had an initial population of 819200 but has been contaminated. Its population is now decreasing by half every hour.

Thus;

After 1 hour, P = 409600

After 2 hours, P = 204800

After 3 hours, P = 102400

Thus;

At t = 1, we have;

409600 = 819200(e^(k × 1))

409600/819200 = e^(k)

e^(k) = 0.5 - - - (eq 1)

At t = 2, we have;

204800 = 819200(e^(k × 2))

204800/819200 = e^(2k)

e^(2k) = 0.25 - - - (eq 2)

To find k, let's divide eq 1 by eq 2.

(e^(2k))/e^(k) = 0.25/0.5

e^(k) = 0.5

k = In 0.5

k = -0.6931

Thus;

P = 819200(e^(-0.6931t))

We want to find the time when the 2 cultures will have the same population. Thus;

200e^(0.6931t) = 819200(e^(-0.6931t))

Arranging, we have;

(e^(0.6931t))/(e^(-0.6931t)) = 819200/200

e^(0.6931t - (-0.6931t) = 4096

e^(1.3862t) = 4096

1.3862t = In 4096

1.3862t = 8.3178

t = 8.3178/1.3862

t ≈ 6 hours