Question 1197652: The growth model for a population of bacteria is given by P(t)=35e^2t where t is measured in days. How many days will it take until the number of bacteria has exceeded 17,500?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
A useful saying: "If the variable is in the trees, then log it down". By "trees", I'm referring to exponents.
We'll use logarithms to isolate the exponent in the variable, so we can isolate the variable itself.
With that in mind, we replace P(t) with 17500 and solve like so.
P(t) = 35e^(2t)
17500 = 35e^(2t)
17500/35 = e^(2t)
500 = e^(2t)
Ln(500) = Ln( e^(2t) ) ... this is where logs come in
Ln(500) = 2t*Ln(e) .... use the rule log(A^B) = B*log(A)
Ln(500) = 2t*1
2t = Ln(500)
t = Ln(500)/2
t = 3.1073040492111
It will take a little over 3 days to reach a population of 17,500 bacteria.
For day 0, aka the starting day, we have:
P(t) = 35e^(2t)
P(0) = 35e^(2*0)
P(0) = 35e^(0)
P(0) = 35*1
P(0) = 35
There are 35 bacteria at the start.
Let's see how many there are for day 1.
P(t) = 35e^(2t)
P(1) = 35e^(2*1)
P(1) = 35e^(2)
P(1) = 35*7.38905609893584
P(1) = 258.616963462754
So about 258 bacteria if we only consider whole numbers of bacteria.
Now day 3
P(t) = 35e^(2t)
P(3) = 35e^(2*3)
P(3) = 35e^(6)
P(3) = 35*403.428793493586
P(3) = 14,120.0077722756
About 14120 bacteria at this point. We're getting pretty close to the target 17,500.
Let's look at day 4
P(t) = 35e^(2t)
P(4) = 35e^(2*4)
P(4) = 35e^(8)
P(4) = 35*2980.9579870501
P(4) = 104,333.529546753
We've reached about 104333 bacteria which is definitely way over the target of 17,500.
These numeric results help show that we reach a population of 17,500 in the interval 3 < t < 4, which helps confirm the answer we reached a while back earlier.
Answer: Approximately 3.107304 days.
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