SOLUTION: dear tutors i have been stumped with this question that did not come from a textbook, please try to answer it for me, thanks. log(2x) + log(x-1)

Question 95610: dear tutors i have been stumped with this question that did not come from a textbook, please try to answer it for me, thanks.
log(2x) + log(x-1) - 2 log(x) = 1. you need to solve for x. i tried converting it into exponential form 10 = 2x(x-1)/x^2, = (2x^2-2x)/x^2 = (2x-2)/x and then multiplication of both sides by x to yield: 10x = 2x-2, then subtraction of 2x from both sides becomes: 8x = -2, division of both sides by 8 yields: x = -1/4 but it is impossible for x to have a negative value since it is impossible to perform a logarithm on a negative? thank you so much for taking the time to figure this out i really appreciate it
Answer by stanbon(48535) (Show Source):
You can put this solution on YOUR website!
log(2x) + log(x-1) - 2 log(x) = 1
Use the law that says loga + logb = logab
and the law nloga = loga^n
and the law loga - logb = log(a/b)
to get:
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log[2x(x-1)]- logx^2 = 1
log [(2x^2-2x)/x^2] = 1
log [(2x-2)/x]= 1
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Now change to exponential form to get:
[(2x-2)/x] = 10^1
2x-2 = 10x
x-1 = 5x
4x = 1-
x=-1/4

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Check this answer in the original equation:
log(2x) + log(x-1) - 2 log(x) = 1
x=-1/4 ?
You get log(negative) + log(negative) - 2log(negative) = 1
But there are no logs of negative values
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Conclusion: No solution
Cheers,
Stan H.

SOLUTION: dear tutors i have been stumped with this question that did not come from a textbook, please try to answer it for me, thanks. log(2x) + log(x-1) - 2 log(x) = 1. you need to solve