Question 22137
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log2x+log2(x-6)=4
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The solution, (x = 8, or x = - 2) by the other person who responded, is PARTIALLY WRONG!!

{{{log (2, (x)) + log (2, (x - 6)) = 4}}}
The SMALLER log argument, x - 6, MUST be > 0. So, x - 6 > 0 ===> x > 6.
We then have: {{{log (2, (x)) + log (2, (x - 6)) = 4}}}, with x > 6.
                                         {{{log (2, x(x - 6)) = 4}}} ----- Applying {{{log (b, (c)) + log (b, d))}}} = {{{log (b, (c*d))}}}
                                                  {{{x(x - 6) = 2^4}}} --- Converting to EXPONENTIAL form <=== Note that the other person has {{{2^4}}} i/o {{{4^2}}},
                                                                                                                                                       but both have the same value, 16
                                                 {{{x^2 - 6x = 16}}} 
                                          {{{x^2 - 6x - 16 = 0}}}
                                       (x - 8)(x + 2) = 0 
                                                      x - 8 = 0     OR    x + 2 = 0 ---- Setting each FACTOR equal to 0
                                                            x = 8     OR           x = - 2

The x-value, 8, is > 6, but - 2 is NOT. This makes - 2 an EXTRANEOUS solution!! So, x = 8 is the only VALID/ACCEPTABLE solution!!</font></font></font></b></pre>