P(t) = P₀ e^(kt)
Where P₀ is the initial population,
P(t) is the population after "t" time.
t is your rate (can be hours, days, years, etc. in this case, hours)
k is the growth constant for this particular problem.
So using the information given, solve for k:
P₀ = 2000
P(4) = 2600
P(t) = P₀ e^(kt)
2600 = 2000e^(k * 4)
1.3 = e^(4k)
Natural log of both sides:
ln(1.3) = 4k
k = ln(1.3) / 4
Now that we have a value for "k", use that, the same P₀, then solve for P(17):
P(t) = P₀ e^(kt)
P(17) = 2000 e^(17ln(1.3) / 4)
Using a calculator to get ln(1.3) then to simplify from there, we get:
P(17) ≈ 2000 e^(17 * 0.262364 / 4)
P(17) ≈ 2000 e^(4.460188 / 4)
P(17) ≈ 2000 e^(1.115047)
P(17) ≈ 2000 * 3.0497
P(17) ≈ 6099.4
Rounded to the nearest unit:
P(17) ≈ 6099 bacteria hope i could help =)))