Question 95610
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.