Question

The number of bacteria in a culture grew from 275 to 1135 in three hours. Find the amount of time needed for the number of bacteria to grow to 5000.

Answer 1

Answer:

6.1 hours

Step-by-step explanation:

The starting number 275 is multiplied by the factor 1135/275 in 3 hours. This gives rise to the exponential model ...

... n(t) = 275·(1135/275)^(t/3)

We want to find t when n(t) = 5000. Substituting this into the equation and solving, we have ...

... 5000 = 275·(1135/275)^(t/3)

... 5000/275 = (1135/275)^(t/3) . . . . divide by 275

... log(5000/275) = (t/3)·log(1135/275) . . . . take the log

... t = 3·log(5000/275)/log(1135/275) . . . . multiply by the inverse of the x-coefficient

... t ≈ 6.13795 . . . hours

Rounded to reasonable precision, the time is approximately 6.1 hours.

Answer 2

Answer:

  6.1 hours

Step-by-step explanation:

Given : The number of bacteria in a culture grew from 275 to 1135 in three hours.

To find : Amount of time needed for the number of bacteria to grow to 5000.

Solution : Let the bacteria undergo through an exponential growth

$y(t)=A e^{kt}$ where,

y(t)= value at time t

A= original value

t=time

k = rate of growth

First, we find the rate of growth in culture grew from 275 to 1135 in three hours.

$y(t)=A e^{kt}$

$1135=275 e^{3k}$

$\frac{1135}{275}=e^{3k}$

$4.13=e^{3k}$

Taking ln(natural log) both side

$ln(4.13)=3k$

$1.42=3k$

$k=0.47$

Now, we find the time needed for the number of bacteria to grow to 5000

$y(t)=275 e^{0.47t}$

$5000=275 e^{0.47t}$

$\frac{5000}{275}=e^{0.47t}$

$18.18=e^{0.47t}$

Taking ln(natural log) both side

$ln(18.18)=0.47t$

$2.900=0.47t$

$\frac{2.900}{0.47}=t$

$t=6.17$

Therefore, Amount of time is approx 6.1 hours