Find the number?

Question

 

The number of bacteria in a culture is growing at a rate of 3,000e^(2t/5) per unit of time t. At t=0, the number of bacteria present was 7,500. Find the number present at t=5.

Answer: 7,500e^2

Answer 1

dN/dt=3000e^(2t/5)

 

N(t)=7500e^(2t/5)+const

 

Since N(0)=7500, const=0

 

N(5)=7500e^2*5/5=7500e²