SOLUTION: log(A)=log(A[0])+0.1tlog(.8) Solve for A as a function of t. Having a bit of difficulty on where to start here. Please help an old man going back to school...

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: log(A)=log(A[0])+0.1tlog(.8) Solve for A as a function of t. Having a bit of difficulty on where to start here. Please help an old man going back to school...      Log On


   

Question 745760: log(A)=log(A[0])+0.1tlog(.8) Solve for A as a function of t.
Having a bit of difficulty on where to start here. Please help an old man going back to school...
Answer by KMST(5396) About Me  (Show Source):
You can put this solution on YOUR website!
I would start with properties of logarithms, like
b%2Alog%28%28a%29%29=log%28%28a%5Eb%29%29 and log%28%28x%29%29%2Blog%28%28y%29%29=log%28%28xy%29%29
log%28%28A%29%29=log%28%28A%5B0%5D%29%29%2B0.1t%2Alog%28%280.8%29%29
log%28%28A%29%29=log%28%28A%5B0%5D%29%29%2Blog%28%280.8%5E%280.1t%29%29%29
log%28%28A%29%29=log%28%28A%5B0%5D%2A0.8%5E%280.1t%29%29%29
A=A%5B0%5D%2A0.8%5E%280.1t%29

EXTRA:
A as a function of t is an exponential decay function.
The graph looks like this
graph%28300%2C300%2C-10%2C90%2C-15%2C135%2C100%2A0.8%5E%280.1x%29%29
If you start with a quantity A%5B0%5D and lose 20% of what you have every 10 hours, at time t=10 hours you will have 80% of A%5B0%5D left
A=A%5B0%5D%2A0.8%5E%280.1%2A10%29=A%5B0%5D%2A0.8%5E1=A%5B0%5D%2A0.8
At time t=20 hours you will have 80% of that 80% left
A=A%5B0%5D%2A0.8%5E%280.1%2A20%29=A%5B0%5D%2A0.8%5E2=A%5B0%5D%2A0.64 , which is 64% of A%5B0%5D
At time t=30 hours you will have 80% of 80% of 80% left
A=A%5B0%5D%2A0.8%5E%280.1%2A30%29=A%5B0%5D%2A0.8%5E3=A%5B0%5D%2A0.512 , which is 51.2% of A%5B0%5D