The count in a bacteria culture was 300 after 10 minutes and 1400 after 40 minutes..?

Question

The count in a bacteria culture was 300 after 10 minutes and 1400 after 40 minutes. Assuming the count grows exponentially,

What was the initial size of the culture?   

Find the doubling period.   

Find the population after 75 minutes.   

When will the population reach 12000? 

Answer 1

The exponential growth equation is in the form of: y(t) = y₀ * e^(k*t), y₀ is the initial size, t is the time, and k is the growth constant.

Using the given information we get

Data 1:

Equation 1 is y(10) = 300 = y₀ * e^(k*10)

Data 2:

Equation 2 is y(40) = 1400 = y₀ * e^(k*10)

We have 2 equations, and 2 unknowns. The easiest approach would be to solve each equation for y_0.

From Equation 1: y₀ = 300 * e^-10k

From Equation 2: y₀ = 1400 * e^-40k

Now we can set the equations equal to each other since they have both been solved for a common variable.

300 * e^-10k = 1400 * e^-40k

Solving for k:

e^-10k / e^-40k = 1400 / 300

e^(30k) = 14/3

ln(e^(30k)) = ln(14/3)

30k = ln(14/3)

k = ln(14/3)/30 ~= 0.051348168

Now k can be used in any of the original equations to solve for y₀

Doing so gives y₀ ~= 179.52

Substituting these results into the exponential growth equation gives:

y(t) = 179.52 * e^{(0.051348168*t)}

The initial size of the culture is always y₀.

The proof is in knowing that the initial size occurs at time = 0.

y(0) = y₀ * e^(k*0)

y(0) = y₀ * e⁽⁰⁾ = y₀ * 1 = y₀

In finding the doubling period you are finding where y(t) = 2 * y₀

The longer method would be to set the equation equal to 2*179.52 and solve for t:

y(t)= 359.04 = 179.52 * e^{(0.051348168*t)}

The more elegant solution is knowing that since

y(t) = 2 * y₀ and y(t) = y₀ * e^(k*t)

then

2 * y₀ = y₀ * e^(k*t)

or better yet

2 = e^(k*t)

Solving for t gets:

t_dbl = 1/k * ln(2)

For this scenario:

t_dbl = 1/0.051348168 * ln(2)

Population at 75 minutes,

y(75) = 179.52 * e^{(0.051348168 * 75)} = 8445.71

What time will the population hit 12,000

Use the long method in the time to double above, and solve for t.

y(t) = 12000 = 179.52 * e^{(0.051348168t)}

t = 1/k * ln(y(t) / y₀) = 1/0.051348168 * ln(12000/179.52) = 81.8 minutes

Hope this helps

Answer 2

300 = a₀e^10k

1400 = a₀e^40k

14 / 3 = e^30k

ln (14/3) = 30k

300 = a₀e^10k substitute value of k and solve for a₀

ln 2 = e^tk substitute value of k and solve for t

A = a₀e^75k substitute values of a₀ and k and solve for A

12000 = a₀e^tk substitute values for a₀ and k and solve for t