The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours it is observed that 300 bacteria are present. After 10 hours 4000 bacteria are present. What was the initial number of bacteria? (Round your answer to the nearest integer.)

Accepted Solution

Answer:137Step-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^kNow 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^10which reduces to:4000^3       300^10------------- = -------------        1             P^7or         1         P^7------------  =  -----------4000^3       300^10or:            300^10P^7 = ---------------- = 9.226*10^13  = 92.26*10^12            4000^3Taking the 7th root of both sides results in P = (92.26*10^12)^(1/7), or                                                                         P = 137.26The initial number of bacteria was 137.