SOLUTION: If something decays according to the equation y=ae^-0.0974t, where t- is in days. After 15 days, 20 grams of the compound remain. What was the original amount of the compound? Pl

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: If something decays according to the equation y=ae^-0.0974t, where t- is in days. After 15 days, 20 grams of the compound remain. What was the original amount of the compound? Pl      Log On


   



Question 1022772: If something decays according to the equation y=ae^-0.0974t, where t- is in days. After 15 days, 20 grams of the compound remain. What was the original amount of the compound? Please show and explain the steps. Thanks
Found 2 solutions by Alan3354, fractalier:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
If something decays according to the equation y=ae^-0.0974t, where t- is in days. After 15 days, 20 grams of the compound remain. What was the original amount of the compound?
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a = amount 15 days earlier.
y = amount now
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y = a*e^(-0.0974*15) = 20
a = 20/e^(-0.0974*15)
a =~ 86.205 grams

Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Okay, from
y=ae^-0.0974t
we plug in for t and y and solve for a...
(Of course we could solve for a first and then plug in...)
20+=+a%2Ae%5E%28-0.0974%2A15%29
Now evaluate the exponential expression and get
20+=+a%2A%28.232%29
so that, when we divide by .232, we get
a = 86.2 grams