Which function can be used to find the number of cells of bacteria in the population at time t

db / dt = kb  

this becomes b(t) = Ce^(kt)  

C = 100, the initial population  

P(1) = 420 = 100 e^(1k)  

4.2 = e^k  

ln 4.2 = k  

a) thus, b(t) = 100 e^(t ln 4.2)  

b) b(3) = 100 e^(3 ln 4.2)  

c) growth constant will still be ln 4.2 (constant percentage of populatioin)  

d) 10000 = 100 e^(t ln 4.2)  

100 = e^(t ln 4.2)  

ln 100 = t ln 4.2  

t = ln 100 / ln 4.2

Step-by-step explanation:

the answer is ! ?

6e14 is the answer.