let the number be n
n = a e^(kt), where a is the initial number and t is the time in hours, k is a constant
case 1, when t=2, n = 125
125 = a e^2k
case 2 , when t = 4, n= 350
350 = a e^4k
divide the two equations
350/125 = a e^4k / (a e^(2k))
2.8 = e^2k
2k = ln 2.8
k = .5148
in first equation,
125 = a e^(2(.5148))
a = 44.64
a) so initially there were 46 bacteria.
b) n = 44.64 e^(.5148t)
c) when t = 8 , n = 44.64 e^(8(.5148)) = appr. 2744
d) 25000 = 44.64 e^(.5148 t)
560.0358 = e^ .5148t
.5148t = ln 560.0358
t = 12.29 hours
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