# Calculus

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 69 cells.
(a) Find the relative growth rate. (Assume t is measured in hours.)
k = ln(8)

(b) Find an expression for the number of cells after t hours.
P(t) = 69*8^t

(c) Find the number of cells after 8 hours.
1157627904
cells

(d) Find the rate of growth after 8 hours. (Round your answer to three decimal places.)
billion cells per hour

(e) When will the population reach 20,000 cells? (Round your answer to two decimal places.)

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1. I suppose you want only the last two parts.

You were told that
N(t) = 69(8^t) is correct
and you got N'(t) = ln(8)(69)(8^t) <----- k(69)(8^t) , so k = ln(8)

so the rate of growth after 8 hrs
= ln(8)*1157627904

e) 69(8^t) = 20000
8^t = 289.85507..
take ln of both sides
t ln(8) = ln(289.85507..)
t = ln(289.85507..) / ln8 = ....

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