Question 166689
<pre>
log3(x-2)+log3(x+4)=3
*********************
<font face = tahoma><font size  = 3><font color = blue><b>
The other person who responded is WRONG!! His solutions, "x = {3.243, -5.243}" are WRONG.
If he'd checked his answers, he would've realized this.

The SMALLER log argument, x - 2 MUST be > 0, so x - 2 > 0, and we get: x > 2. We now have:
{{{log (3, (x - 2)) + log (3, (x + 4)) = 3}}}, with x being > 2.
        {{{log (3, (x - 2)(x + 4)) = 3}}} ----- Applying {{{log (b, (c)) + log (b, (d))}}} = {{{log (b, (c*d))}}}
              {{{(x - 2)(x + 4) = 3^3}}} ---- Converting to EXPONENTIAL form
                 {{{x^2 + 2x - 8 = 27}}}
           {{{x^2 + 2x - 8 - 27 = 0)}}}
               {{{x^2 + 2x - 35 = 0}}}
       (x - 5)(x + 7) = 0
        x - 5 = 0        OR        x + 7 = 0
              x = 5        OR               x = - 7 (ignore)

The constraint above "states" that x MUST be > 2. 5 is > 2, but - 7 is NOT. This makes - 7 an
EXTRANEOUS solution, and the only ACCEPTABLE solution, x  = 5.

You can do the CHECK!!</pre>