F(t) = 30b^6 = 300,
30b^6 = 300,
Divide both sides by 30:
b^6 = 10,
Take log of both sides:
6log(b) = log10,
log(b) = log10 / 6,
log(b) = 1/6,
Exponential form:
10^(1/6) = b,
Or b = 10^(1/6).
F(t) = 30*10^(1/6)^t,
a. Eq: F(t) = 30*10^(t/6).
b. F(14)=30*10^(14/6) = 6463 Infected.
c. 30*10^(t/6) = 10000,
Divide both sides by 30:
10^(t/6) = 10000/30.
(t/6)log10 = log10000 - log30 = 2.5229.
(t/6)log10 = 2.5229.
Divide both sides by log10:
t/6 = 2.5229,
Multiply both sides by 6:
t = 6 * 2.5229 = 15.14 = approximately 15 Days.
PLEASE HELP WITH THIS-ITS DUE APRIL 5TH, 2011!!!
During a past epidemic, the number of infected people increased exponentially with time, or f (t) = ab^t , where f is the number of infected people at time t, in days. Suppose that at the onset of the epidemic (0 days) there are 30 people infected, and 6 days after the onset there are 300 people infected.
a) Specify the function f, that models this situation, by determining a and b.
b) Predict the number of people who will be infected at the end of 2 weeks using your answer to a.
c) According to this function, approximately when will the number of infected people reach 10,000? (Use logarithms to sove this part - a guess and check answer WILL NOT be accepted)
1 answer