SOLUTION: The number of bacteria in a certain population increases according to a continuous exponential growth model, with a growth rate parameter of
8.7% per hour. How many hours does it
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8.7% per hour. How many hours does it
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Question 1065322: The number of bacteria in a certain population increases according to a continuous exponential growth model, with a growth rate parameter of
8.7% per hour. How many hours does it take for the size of the sample to double? Answer by Theo(13342) (Show Source):
f is the future value
p is the present value
e is the scientific constant of 2.718281828.....
r is the rate of growth per time period.
t is the number of time periods.
in your problem:
f = 2
p = 1
r = .087 per hour.
t = number of hours.
formula becomes 2 = 1 * e^(.087 * t)
take the natural log of both sides of this equation to get:
ln(2) = ln(1 * e^(.087 * t)
this becomes ln(2) = ln(1) + .087 * t * ln(e)
since ln(1) = 0 and ln(e) = 1, this becomes ln(2) = .087 * t
solve for t to get t = ln(2) / .087 = 7.967208972
the sample should double in 7.967208972 hours.
replace t with 7.967208972 in your original equaiton and you get:
2 = 1 * e^(.087 * 7.967208972).
evaluate this equation to get 2 = 2
this confirms the solution is correct.
note that:
ln(1) = 0
ln(e) = 1
ln(a * b^c) = ln(a) + ln(b^c) which is then equal to ln(a) + c*ln(b)
you can use your calculator to confirm the first 2 statements are true.
the third statement is based on the properties of logarithms.
