Question 416851
<pre>
log base 2 (x-2)+log base 2(8-x)-log base 2(x-5)=3

{{{log (2, (x - 2)) + log (2, (8 - x)) - log (2, (x - 5)) = 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 < {{{(- 8)/(- 1)}}}, so x < 8. We now get: 
{{{log (2, (x - 2)) + log (2, (8 - x)) - log (2, (x - 5)) = 3}}}, <font color = red><font size = 4><b>with 5 < x < 8 </font></font></b>

            {{{log (2, (((x - 2)(8 - x))/(x - 5))) = 3}}}
                  {{{((x - 2)(8 - x))/(x - 5) = 2^3}}}
                  {{{((x - 2)(8 - x))/(x - 5) = 8}}}
                 (x - 2)(8 - x) = 8(x - 5)
                 {{{8x - x^2 - 16 + 2x = 8x - 40}}}
          {{{- x^2 + 10x - 8x - 16 + 40 = 0}}}
                   {{{- x^2 + 2x + 24 = 0}}} <=== As seen HERE, the quadratic equation is NOT x^2 - 18x - 24 = 0, as @Stanbon states. 
               {{{- x^2 + 6x - 4x + 24 = 0}}} 
          - x(x - 6) - 4(x - 6) = 0
               (x - 6)(- x - 4) = 0
                x - 6 = 0       OR    - x - 4 = 0
                   <font color = red><font size = 4><b>x = 6</font></font></b>       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!!</pre>