can someone plz help me with this math problem?

Question

The number of bacteria in a culture is growing at a rate of 3000e^(2t/5) per unit of time at t=0 the number of bacteria present was 7500. find the number present at t=5

Answer 1

Your formula is incorrect. At t=0 the number of bacteria would be 3000*e^(2*0/5) =3000*e^0 = 3000*1 = 3000 bacteria and not 7500 as you say......????

Answer 2

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

Integrating gives :

N = 3000e^(2t/5) * 5/2 + C, where C is a constant.
= 7500e^(2t/5) + C

At t = 0, 7500 = 7500e^(2*0/5) + C
Therefore, C = 7500 -7500 = 0

The equation is thus: N = 7500e^(2t/5)

At t = 5, N = 7500e^(2*5/5) = 7500e^2 = 55418.