Question 327293
A certain population obeys the natural law of decay. at a certian time,the population is 40,000. three years later the population is 38000. approximately how long does it take the population to decline from 40,000 to 20,000?
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You have two points (0,40000) and (3,38000)
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Assuming the decay is exponential, use the 
model A(t) = ab^t
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Solve for a and b
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Using (0,40000), solve for "a":
40,000 = ab^0
a = 40,000
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Now, A(t) = 40,000*b^t
Using (3,38000), solve for "b":
38000, = 40,000*b^3
19/20 = b^3
b = 0.983
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Equation:
A(t) = 40000*0.983^t
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When does the population decay to 20,000?
20,000 = 40000*0.983^t
0.983^t = 0.5
t = log(0.5)/log(0.983)
t is approximately 40
Rounded up you would get 41 years
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Cheers,
Stan H.