SOLUTION: I know the answer is no solution but i need to know how to work this problem please help {{{6+log(2,(2x-4))=7+log(2,(3x))

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I know the answer is no solution but i need to know how to work this problem please help {{{6+log(2,(2x-4))=7+log(2,(3x))      Log On


   



Question 64567: I know the answer is no solution but i need to know how to work this problem please help
6+log(2,(2x-4))=7+log(2,(3x))

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
6+log(2,(2x-4))=7+log(2,(3x))
Keep in mind, the logs are base two.
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log(2x-4)=1+log(3x)
log(2x-4)-log(3x)=1
log[(2x-4)/3x]=1
Since the logs are base 2 you get:
(2x-4)/3x=2^1
2x-4=6x
4x=-4
x=-1
Checking this in the original equation you get log(-3) which doesn't exist.
So, no solution.
Cheers,
Stan H.