Question

If the number of bacteria in a colony doubles every 72 hours and there is currently a population of 105,700 bacteria, what will the population be 216 hours from now? bacteria

Answer 1

EXPLANATION

Let's see the facts:

Period = 72 hours

Initial Population= 105,700

Time = 216 hours

The equation is as follows:

`P=P_0e^(rt)^{}`

P_0=Initial Population

r= rate of growth

t=time

First, we need to find r:

In 72 hours ------> 2*105,700 = 211,400 bacteria

Replacing terms:

`211,400=105,700e^((r72))`

Dividing both sides by 105,700:

`(211,400)/(105,700)=e^((72r))`

Applying ln to both sides:

`\ln ((211,400)/(105,700))=\ln (e^(72r))`

Simplifying:

`\ln ((211,400)/(105,700))=72r\cdot\ln e`

Dividing both sides by 72:

`(\ln ((211,400)/(105,700)))/(72)=r`

Switching sides:

`r=(\ln ((2114)/(1057)))/(72)`

Simplifying:

`r=(0.69)/(72)=0.009627`

Now that we have r=0.009627 we can calculate the value of P as shown as follows:

`P=105,700e^((0.009627\cdot216))`

Multiplying terms:

`P=105,700\cdot e^((2.08))^{}`

Now, we can solve the expression:

`P=105,700\cdot8=845,600`

The answer is 845,600 bacteria.