In a certain culture of bacteria, the rate of increase is proportional to the number present. If there are 10000 at the end of 3hours and 40000 at the end of 5hours,how many were there in yhe beginning?

Answer 1

so number = a e^(kt)
10,000 = a e^(3k)
40,000 = a e^(5k)
divide the 2nd by the 1st:
4 = e^(2k)
2k = ln4
k = (ln4)/2

so N = a e^( (ln4)/2 t)

so we need a

a e^(3ln4/2) = 10,000
a = 10,000/8 = 1250
initial count is 1250 bacteria

Answer 2

so number = a e^(kt)
10,000 = a e^(3k)
40,000 = a e^(5k)
divide the 2nd by the 1st:
4 = e^(2k)
2k = ln4
k = (ln4)/2

so N = a e^( (ln4)/2 t)

so we need a

a e^(3ln4/2) = 10,000
a = 10,000/8 = 1250
initial count is 1250 bacteria