Question

A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 hours, there are 10,000 bacteria. At the end of 5 hours, there are 40,000 bacteria. How many bacteria were present initially?

Answer 1

Answer: There were 1,250 bacteria present initially.

Step-by-step explanation:

To find the exponential growth to this, lets create an exponential model shown as below:

f(x) = a($b^{y}$)

 Let y represent the amount of hours have passed after the third-hour recording. With the information given, we can create a set of input-output pairs: (0 , 10000) and (2 , 40000). Now that we have this set of pairs, we have given ourselves the initial value of the function, a = 10,000. We can now substitute the second point into the equation using N = 40,000.

N(y) = 10000b^y

40000 = 10000b^2    Divide and write in lowest terms.

÷10000   ÷10000

4 = b^2

b = $(4)^{\frac{1}{2} }$ Isolate b using properties of exponents.

b = 2

Now that we have the exponential growth, we can use b = 2 to work backwards and find the initial amount of bacteria.

10,000 ÷ 2 = 5,000 <-- Amount of bacteria 2nd hour

5,000 ÷ 2 = 2,500 <-- Amount of bacteria 1st hour

2,500 ÷ 2 = 1,250 <-- Initial amount of bacteria.

The initial amount of bacteria is 1,250