SOLUTION: A bacteria colony grows continuously at an exponential rate. There are initially 1.6 million present. After 4 days, there are 6.6 million present. a. What is an equation modeling

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Question 1005174: A bacteria colony grows continuously at an exponential rate. There are initially 1.6 million present. After 4 days, there are 6.6 million present.
a. What is an equation modeling the number of bacteria present after "d" days?
b. How many bacteria will be present after 9 days?
c. How long will it take the number of bacteria to reach 90 million?
Can someone explain this in detail? I am really having trouble with the question.
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
N=Noexp(kt)
6.6=1.6exp(4k)
4.125=exp(4k)
ln of both sides
4k=1.4170.
without rounding k=0.3543 (now round)
1.6*e(9*.3543)=38.799 million
90/1.6=exp(0.3543t)
17.45=0.3543t,
ln of both sides
8.070 days.
The first thing you need to do is set up the exponential equation with some constant k of exponentiation. Then solve for the constant using the data. You have to take ln of both sides, because that removes the exp part. You then look for the other variable, time, once you know what the constant is.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A bacteria colony grows continuously at an exponential rate. There are initially 1.6 million present. After 4 days, there are 6.6 million present.
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y = ab^x
a = 1.6 million
6.6 million = 1.6*b^4
b^4 = 4.125
b = 1.425
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a. What is an equation modeling the number of bacteria present after "d" days?
y = 1.6*1.425^d
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b. How many bacteria will be present after 9 days?
y = 1.6*1.425^9 = 38.77 million
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c. How long will it take the number of bacteria to reach 90 million?
90 = 1.6*1.425^x
1.425^x = 56.25
x = log(56.25)/log(1.425)
x = 11.38 days
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Cheers,
Stan H.
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