SOLUTION: Solve the equation if possible. 12e^8t-e=8e^8t I don't really know how to go about solving this. Do I subtract the 12e^8t to the other side first, or something else? Also, it's

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Solve the equation if possible. 12e^8t-e=8e^8t I don't really know how to go about solving this. Do I subtract the 12e^8t to the other side first, or something else? Also, it's      Log On


   

Question 278941: Solve the equation if possible.
12e^8t-e=8e^8t
I don't really know how to go about solving this. Do I subtract the 12e^8t to the other side first, or something else? Also, it's possible that the answer doesn't exist or that the exact answer doesn't exist. Please help!
Found 3 solutions by Alan3354, Edwin McCravy, solver91311:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
12e%5E8t-e=8e%5E8t
Subtract 8e^8t
4e%5E8t+-+e+=+0
4e%5E8t+=+e
4e%5E%288t-1%29+=+1
e%5E%288t-1%29+=+1%2F4
8t-1 = ln(1/4)
8t = 1 + ln(1/4)
t = (1 + ln(1/4))/8
t =~-0.048286795

Answer by Edwin McCravy(20062) About Me  (Show Source):
You can put this solution on YOUR website!

12e%5E%288t%29-e=8e%5E%288t%29

12e%5E%288t%29-8e%5E%288t%29=e

4e%5E%288t%29=e


Divide both sides by e

%284e%5E%288t%29%29%2Fe=e%2Fe

Notice that e is raised to the first power on the
left bottom, so you subtract exponents:

%284e%5E%288t%29%29%2Fe%5E1=1

4e%2A%288t-1%29=1

Take natural logs of both sides:

ln%284e%2A%288t-1%29%29=ln%281%29

Use laws of logarithms:

ln%284%29%2Bln%28e%5E%288t-1%29%29=0

Use another law of logarithms:

ln%284%29%2B%288t-1%29=0

ln%284%29%2B8t-1=0

8t=1-ln%284%29

Divide both sides by 8

t=%281-ln%284%29%29%2F8 or about -.0482867951

Edwin

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!




Yes, the first step is to put both of the terms on one side. Just for the sake of what I consider to be neatness, I'm going to add to both sides, with the following result:



Next, take the natural log of both sides:



The log of the product is the sum of the logs, so:



The log of a base raised to a power is the power times the log of the base:



Since we know that



Add to both sides:



Multiply by



John