Calculus Math High Level Question!!!!! (Again)

Hello, I also need help with this question that is driving me mad, Thanks again!


An experiment is carried out in which the number "n" of bacteria in a liquid, is given by the formula
n = 650e^kt (e to the power of k times t) , where t is the time in minutes after the beginning of the experiment and k is a constant.
The number of bacteria doubles every 20 minutes.

Find

(a) the exact value of k;
(b) the rate at which the number of bacteria is increasing when t = 90 .

a)
When t=20,

2 = e^k(20)
e^20k = 2
20k = ln(2)
k = ln(2)/20 = 0.034657359

n = 650e^0.034657359t
dn/dt = 650 e^0.034657359 t (0.034657359)
Substitute t=90
dn/dt = 650 (0.034657359) e^0.034657359(90) = 509.73
=510 per mninute