SOLUTION: Solve log(x-1)- log6 = log(x-2)- logx

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Question 105760: Solve log(x-1)- log6 = log(x-2)- logx
Found 2 solutions by HyperBrain, MathTherapy:
Answer by HyperBrain(694) About Me  (Show Source):
You can put this solution on YOUR website!
log%28x-1%29-+log%286%29+=+log%28x-2%29-+log%28x%29
log%28%28x-1%29%2F%286%29%29=log%28%28x-2%29%2F%28x%29%29
%28x-1%29%2F%286%29=%28x-2%29%2F%28x%29


By cross multiplication,
6%28x%2B2%29=x%28x-1%29
6x%2B12=x%5E2-x
x%5E2-x=6x%2B12
x%5E2-x%2Bx=6x%2Bx%2B12
x%5E2=7x%2B12
x%5E2-7x=7x-7x%2B12
x%5E2-7x=12
x%5E2-7x-12=12-12
x%5E2-7x-12=0

By the quadratic formula,
x=%287%2B-sqrt%287%5E2-4%2A1%2A%28-12%29%29%29%2F%282%2A1%29
x=%287%2B-sqrt%2849%2B48%29%29%2F%282%29
x=%287%2B-sqrt%2897%29%29%2F%282%29
x=%287%2B+sqrt%2897%29%29%2F%282%29 or x=%287-+sqrt%2897%29%29%2F%282%29

Power up,
HyperBrain!

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Solve log(x-1)- log6 = log(x-2)- logx
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@HyperBrain(694)'s answer, below, is WRONG, and DOUBLE-WRONG for the negative value of x, since x MUST BE > 2.
"x=%287%2B+sqrt%2897%29%29%2F%282%29 or x=%287-+sqrt%2897%29%29%2F%282%29"
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The smallest log variable-argument, x - 2 MUST BE > 0, so x > 2. We then get:
log+%28%28x+-+1%29%29+-+log+%28%286%29%29+=+log+%28%28x+-+2%29%29+-+log+%28%28x%29%29, with x > 2.
          log+%28%28%28x+-+1%29%2F6%29%29+=+log+%28%28%28x+-+2%29%2Fx%29%29 ----- Applying log+%28b%2C+%28c%29%29+-+log+%28b%2C+%28d%29%29 = log+%28b%2C+%28c%2Fd%29%29
                     %28x+-+1%29%2F6+=+%28x+-+2%29%2Fx ----- Applying e = f, since log+%28b%2C+%28e%29%29 = log+%28b%2C+%28f%29%29
                  x(x - 1) = 6(x - 2) --- Cross-multiplying
                     x%5E2+-+x+=+6x+-+12
      x%5E2+-+x+-+6x+%2B+12+=+0
            x%5E2+-+7x+%2B+12+=+0
          (x - 4)(x - 3) = 0
           x - 4 = 0       or       x - 3 = 0 ---- Setting each factor equal to 0
                 x = 4       or            x = 3

Both values of x are VALID/ACCEPTABLE, since both are > 2.