6.4. Exponential Growth and Decay http://www.ck12.org
Example 2:
A certain population of bacteria increases continuously at a rate that is proportional to its present number. The initial
population of the bacterial culture is 140 and jumped to 720 bacteria in 4 hours.
- How many will be there in 10 hours?
- How long will it take the population to double?
Solution:
From reading the first sentence in the problem, we learn that the bacteria is increasing exponentially. Therefore, the
exponential growth formula is the correct model to use.
- Just like we did in the previous example, we need to first findk,the growth rate. Notice thatC= 140 ,t= 4 ,and
y= 720 .Substituting and solving fork.
y=Cekt
720 = 140 ek(^4 ).Dividing both sides by 140 and then projecting the natural logarithm on both sides,
ln^720140 =lne^4 k
ln 5. 143 = 4 k
k= 0. 409.Now that we have foundk,we want to know how many will be thereafter 10 hours. Substituting,
y=Cekt
= 140 e(^0.^409 )(^10 )
=8364 bacteria.- We are looking for the time required for the population to double. This means that we are looking for the time at
whichy= 2 C.Substituting,
y=Cekt
2 C=Cekt
2 =ekt.Solving fortrequires taking the natural logarithm of both sides:
ln 2=lnekt
ln 2=kt.Solving fort,