Question 734837
<pre>
solve log3(x-3) + log3(x+2) = log3 6
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The other person's statement that if "all log bases are the same (namely 3)", then: x-3 + x+2 = 6
                                                                                    2x - 1 = 6 and
                                                                                    2x = 7 and x = 3.5, is a FALLACY!
He couldn't be more WRONG!! If he'd checked, he would've seen that 3.5 doesn't: make sense
                                                                                make the equation TRUE!

The smaller log argument, x - 3 "tells" one that x - 3 MUST be > 0. So, x - 3 > 0____x > 3
We then get: {{{log (3, (x - 3)) + log (3, (x + 2)) = log (3, (6))}}}, with X > 3
                  {{{log (3, (x - 3)(x + 2)) = log (3, (6))}}} ----- Applying {{{log (b, (c)) + log (b, (d))}}} = {{{log (b, (c*d))}}}
                    (x - 3)(x + 2) = 6 ------ Applying c = d, if {{{log (b, (c)) = log (b, (d))}}}
                          {{{x^2 - x - 6 = 6}}}
                       {{{x^2 - x - 6 - 6 = 0}}}
                         {{{x^2 - x - 12 = 0}}}
                    (x - 4)(x + 3) = 0
                     x - 4 = 0      OR     x + 3 = 0    
                         x = 4      OR     x = - 3

As stated above, x MUST be > 3, so ONLY x = 4 is ACCEPTABLE. This makes x = - 3, an EXTRANEOUS solution to this
equation, and is therefore IGNORED/REJECTED.</pre>