Question 747802
<pre>
log3(x+6)+log3(x-6)-log3x=2
solve the logarithmic equation
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It's heart-wrenching to see how the other person (@MATHLOVER) makes this problem so long and time-consuming! Why??
In my opinion, she should learn how to solve problems much more efficiently.
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{{{log (3, (x + 6)) + log (3, (x - 6)) - log (3, (x)) = 2}}}
This log equation starts with 3 log arguments: a1 (x + 6), a2 (x - 6), and a3 (x), the smallest being log argument
a2, or x - 6. Argument "a2" "tells" one that x - 6 MUST be > 0. So, x - 6 > 0_____x > 6 
We now have: {{{log (3, (x + 6)) + log (3, (x - 6)) - log (3, (x)) = 2}}}, with x > 6           
                       {{{log (3, (((x + 6)(x - 6))/(x))) = 2}}} ---- Applying {{{log (b, (a1)) + log (b, (a2)) - log (b, (a3))}}} = {{{log (b, ((a1*a2))/a3)}}}

                               {{{(x + 6)(x - 6)/x = 3^2}}} --- Applying {{{c = b^d}}}, when: {{{log (b, (c)) = d}}}
                               {{{(x + 6)(x - 6)/x = 9}}}
                                  {{{(x^2 - 36)/x = 9}}}
                                    {{{x^2 - 36 = 9x}}} --- Cross-multiplying
                                {{{x^2 - 9x - 36 = 0}}}
                           (x - 12)(x + 3) = 0
                            x - 12 = 0       OR      x + 3 = 0
                                 x = 12      OR          x = - 3 

As stated above, x MUST be > 6, so ONLY x = 12 is ACCEPTABLE. This makes x = - 3, an EXTRANEOUS solution to this
equation, and is therefore IGNORED/REJECTED.</pre>