SOLUTION: Under the onslaught of the college algebra second period class, a pile of homework problems decreased exponentially. It decreased form 900 to 500 problems in only 40 minutes. How l

Question 753000: Under the onslaught of the college algebra second period class, a pile of homework problems decreased exponentially. It decreased form 900 to 500 problems in only 40 minutes. How long would it take until only 300 problems remained?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Under the onslaught of the college algebra second period class, a pile of homework problems decreased exponentially. It decreased form 900 to 500 problems in only 40 minutes. How long would it take until only 300 problems remained?
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y = ab^x
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You have 2 points (0,900) and (40,500)
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Using (0,900) 900 = ab^0
900 = a
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y = 900*b^x
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Using (40,500) solve for "b":
500 = 900*b^40
b^40 = 5/9
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Equation:
f(x) = 900(5/9)^(x/40)
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How long would it take until only 300 problems remained?
Solve: 300 = 900(5/9)^(x/40)
(5/9)^(x/40) = 1/3
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(x/40) = log(1/3)/log(5/9)
x/40 = 1.87
x = 75 minutes
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Cheers,
Stan H.
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