SOLUTION: A rodent population has moved into a new commercial development area in the city.The population initially had 16 animals and displayed an EXPONENTIAL rate of growth.After 486 weeks

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Question 135307: A rodent population has moved into a new commercial development area in the city.The population initially had 16 animals and displayed an EXPONENTIAL rate of growth.After 486 weeks, the population numbered 8192 rodents. What is the doubling time of this rodent population? GROSS :)
Found 2 solutions by vleith, stanbon:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Given exponential growth. N+=+N%5B0%5De%5E%28kt%29
8192+=+16e%5E%28k486%29
512+=+e%5E486k
ln%28512%29+=+486k
6.238+=+486k
0.0128+=+k
So use k to double.
32+=+16e%5E%280.0128t%29
2+=+e%5E%280.0128t%29
ln%282%29+=+0.0128t
0.693+=+0l28t
54.15 weeks = t

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A rodent population has moved into a new commercial development area in the city.The population initially had 16 animals and displayed an EXPONENTIAL rate of growth.After 486 weeks, the population numbered 8192 rodents. What is the doubling time of this rodent population?
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You have two points: (0,16) and (486,8192)
8192 = ab^(486)
16 = ab^0
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From 2nd equation : a = 16
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Divide the 1st equation by the 2nd to get:
512 = b^486
b = 1.0129
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EQUATION:
# of rodents = 16*(1.0129)^x
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Find the power of 2 that equals 1.0129
Solve log2 1.0129 = k
k = log(1.0129)/log2 = 0.018491749...
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EQUATIOn:
# of rodents = 16[(2)^(0.018491749...)^x
or
# of rodents = 16(2)^(x/54.0781728...) where x is number of weeks since there were 16
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Conclusion: The doubling rate is approximately 54.0792 weeks
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Cheers,
Stan H.