Answer:
137
Step-by-step explanation:
Use the model A = Pe^(kt), where P is the initial number and k is the growth constant. Then, in case 1,
300 = Pe^(k*3)
and in case 2,
4000 = Pe^(k*10). We need to find P and e^k.
300 = Pe^(k*3) can be rewritten as (300/P) = [e^k]^3, which in turn may be solved for e^k:
∛(300/P) = e^k
Now we go back to 4000 = Pe^(k*10) and rewrite it as (4000/P) = [e^k]^10. Substituting ∛(300/P) for e^k, we obtain:
(4000/P) = (300/P )^(10/3)
which must now be solved for P. Raising both sides by the power of 3, we get:
(4000/P)^3 = (300/P)^10. Therefore,
4000^3 300^10
------------- = -------------
P^3 P^10
which reduces to:
4000^3 300^10
------------- = -------------
1 P^7
or
1 P^7
------------ = -----------
4000^3 300^10
or:
300^10
P^7 = ---------------- = 9.226*10^13 = 92.26*10^12
4000^3
Taking the 7th root of both sides results in P = (92.26*10^12)^(1/7), or
P = 137.26
The initial number of bacteria was 137.
