SOLUTION: At time t = 0 hours, there are 500 bacteria in a favorable growth medium. 5 hours later, there are 2000 bacteria. Assuming exponential growth, what is the growth constant "k" for t
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-> SOLUTION: At time t = 0 hours, there are 500 bacteria in a favorable growth medium. 5 hours later, there are 2000 bacteria. Assuming exponential growth, what is the growth constant "k" for t
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Question 1199252: At time t = 0 hours, there are 500 bacteria in a favorable growth medium. 5 hours later, there are 2000 bacteria. Assuming exponential growth, what is the growth constant "k" for the bacteria?
A. 0.155
B. 2.543
C. 0.135
D. 0.277
E. 1.467
F. 1.866 Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! it looks like they used the continuous compounding formula of f = p * e ^ (rt)
f is the future value = 2000
p is the present value = 500
r is the interest rate per hour
t is the number of hours = 5
formula becomes:
formula becomes 2000 = 500 * e ^ (r * 5)
divide both sides of the equation by 4 to get:
2000 / 500 = ^ (r * 5)
simplify to get:
4 = e ^ (5r)
take the natural log of both sides of the equation to get:
ln(4) = ln(e ^ (5r))
this becomes ln(4) = 5r * ln(e) which becomes ln(e ^ 5r)) = 5r * ln(e)
this then becomes ln(4) = 5r because ln(e) = 1
divide both sides of the equation by 5 to get ln(4) / 5 = r
solve for r to get r = .277 rounded to 3 decimal places.
looks like selection D is your answer.
