Calculus II Exponential Growth and Decay

Question

A certain bacteria population P obeys the exponential growth law: P(t)=1500e^2.1t (t in hours)

a) How many bacteria are present initially?

b) At what time will there be 10,000 bacteria?

Answer 1

P(t) = 1500e^2.1t

A) At t = 0, P =1500 as e⁰=1

B) P/1500 = e^2.1t

Ln (10000/1500) = 2.1t

1.897 = 2.1t

t = 1.897/2.1 = 0.903 hours