SOLUTION: The amount of water remaining after t days is given by A=A0e^-0.18 where A0 is the starting amount at the beginning of the time period. How much should be acquired now to have 40

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Question 65741: The amount of water remaining after t days is given by A=A0e^-0.18 where A0 is the starting amount at the beginning of the time period. How much should be acquired now to have 40 grams remaining after 3 days?
Found 2 solutions by josmiceli, ankor@dixie-net.com:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
A+=+A%5B0%5De%5E%28-.18t%29
40+=+A%5B0%5De%5E%28-.18%2A3%29
A%5B0%5D+=+40%2A%281%2Fe%5E%28-.54%29%29
ln%28A%5B0%5D%29+=+ln%2840%29+%2B+.54
ln%28A%5B0%5D%29+=+3.689+%2B+.54
ln%28A%5B0%5D%29+=+4.229+
A%5B0%5D+=+68.649 grams answer

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The amount of water remaining after t days is given by A=A0e^-0.18 where A0 is the starting amount at the beginning of the time period. How much should be acquired now to have 40 grams remaining after 3 days?
:
The formula A = Ao*e^-.18 seems to be missing the time factor t:
:
I assume it is A = Ao * e^-.18t
:
Substitute for A and t:
40 = Ao * e^(-.18*3)
40 = Ao * e^-.54
:
Find e^-.54 on a good calc:
40 = Ao * .582748
:
Ao = 40/.582748
Ao = 68.64 grams is the initial amt