Question 105760
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Solve log(x-1)- log6 = log(x-2)- logx
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@HyperBrain(694)'s answer, below, is WRONG, and DOUBLE-WRONG for the negative value of x, since x MUST BE > 2.
"{{{x=(7+ sqrt(97))/(2)}}} or {{{x=(7- sqrt(97))/(2)}}}"
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The smallest log variable-argument, x - 2 MUST BE > 0, so x > 2. We then get:
{{{log ((x - 1)) - log ((6)) = log ((x - 2)) - log ((x))}}}, with x > 2.
          {{{log (((x - 1)/6)) = log (((x - 2)/x))}}} ----- Applying {{{log (b, (c)) - log (b, (d))}}} = {{{log (b, (c/d))}}}
                     {{{(x - 1)/6 = (x - 2)/x}}} ----- Applying e = f, since {{{log (b, (e))}}} = {{{log (b, (f))}}}
                  x(x - 1) = 6(x - 2) --- Cross-multiplying
                     {{{x^2 - x = 6x - 12}}}
      {{{x^2 - x - 6x + 12 = 0}}}
            {{{x^2 - 7x + 12 = 0}}}
          (x - 4)(x - 3) = 0
           x - 4 = 0       or       x - 3 = 0 ---- Setting each factor equal to 0
                 x = 4       or            x = 3

Both values of x are VALID/ACCEPTABLE, since both are > 2.</font></font></font></b></pre>