SOLUTION: log base 2 (x-2)+log base 2(8-x)-log base 2(x-5)=3

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Question 416851: log base 2 (x-2)+log base 2(8-x)-log base 2(x-5)=3
Found 2 solutions by stanbon, MathTherapy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log base 2 (x-2)+log base 2(8-x)-log base 2(x-5)=3
----
log2[(x-2)(8-x)/(x-5)] = 3
----
[(x-2)(8-x)/(x-5)] = 8
(x-2)(8-x) = 8(x-5)
-x^2+10x-16 = 8x-40
x^2-18x-24 = 0
---
Solve using the Quadratic Formula.
Discard values of "x" which cause
the original equation to have log2(negative number).
======================
Cheers,
Stan H.

Answer by MathTherapy(10699) About Me  (Show Source):
You can put this solution on YOUR website!
log base 2 (x-2)+log base 2(8-x)-log base 2(x-5)=3

log+%282%2C+%28x+-+2%29%29+%2B+log+%282%2C+%288+-+x%29%29+-+log+%282%2C+%28x+-+5%29%29+=+3
 Looking at the above, we see that x - 5 is SMALLER than x - 2, so we'll have: x - 5 > 0, which means that x > 5. 
 Also, 8 - x MUST be > 0, so - x > - 8, and x < %28-+8%29%2F%28-+1%29, so x < 8. We now get: 
log+%282%2C+%28x+-+2%29%29+%2B+log+%282%2C+%288+-+x%29%29+-+log+%282%2C+%28x+-+5%29%29+=+3, with 5 < x < 8 

            log+%282%2C+%28%28%28x+-+2%29%288+-+x%29%29%2F%28x+-+5%29%29%29+=+3
                  %28%28x+-+2%29%288+-+x%29%29%2F%28x+-+5%29+=+2%5E3
                  %28%28x+-+2%29%288+-+x%29%29%2F%28x+-+5%29+=+8
                 (x - 2)(8 - x) = 8(x - 5)
                 8x+-+x%5E2+-+16+%2B+2x+=+8x+-+40
          -+x%5E2+%2B+10x+-+8x+-+16+%2B+40+=+0
                   -+x%5E2+%2B+2x+%2B+24+=+0 <=== As seen HERE, the quadratic equation is NOT x^2 - 18x - 24 = 0, as @Stanbon states. 
               -+x%5E2+%2B+6x+-+4x+%2B+24+=+0 
          - x(x - 6) - 4(x - 6) = 0
               (x - 6)(- x - 4) = 0
                x - 6 = 0       OR    - x - 4 = 0
                   x = 6       OR    - 4 = x (IGNORE)

Of the 2 solutions above, ONLY the value 6, for x, is > 5, but < 8. This is why x = 6 is ACCEPTED as a solution, while x = - 4
is IGNORED/REJECTED!!