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)

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

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(75887) (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:
-----------
log[2x(x-1)]- logx^2 = 1
log [(2x^2-2x)/x^2] = 1
log [(2x-2)/x]= 1
--------
Now change to exponential form to get:
[(2x-2)/x] = 10^1
2x-2 = 10x
x-1 = 5x
4x = 1-
x=-1/4

=================
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
-----------
Conclusion: No solution
Cheers,
Stan H.

RELATED QUESTIONS
Dear Tutors, I just wanted to thank each of you who helped me lately. I am awaiting my... (answered by edjones)

Tutor Please Help...This question is from a textbook and I keep getting different... (answered by Earlsdon,rapaljer)

Dear yenigelabert@yahoo.com, your question for free tutors has been submitted. Note, (answered by mananth)

Dear Queenholly@comcast.net, your question for free tutors has been submitted. If... (answered by Alan3354)

Dear MathTutor, I appreciate your response on the question I asked regarding the... (answered by math_tutor2020)

Dear Tutors, I have been trying several different ways to solve this problem but have... (answered by ichudov)

Dear Tutor, Please help me with this CI question...Thanks in advance. Q. Why is a 99% (answered by stanbon)

Dear tutors, I am not really sure if this is under the right context so forgive me if it... (answered by richwmiller)

Thank you tutors for being here to assist those of us who struggle from time to time in... (answered by rapaljer,palanisamy)