Question 685646
The trick is to make the substitution:
{{{ 1 = log( 9,9 ) }}}
{{{  log(9,(x-6)) + log(9,(x-6)) = 1 }}}
{{{  log(9,(x-6)) + log(9,(x-6)) = log( 9,9 ) }}}
{{{  log(9,(x-6)^2) = log( 9,9 ) }}}
{{{ ( x-6 )^2 = 9 }}}
{{{ x^2 - 12x + 36 = 9 }}}
{{{ x^2 - 12x + 27 = 0 }}}
{{{ ( x - 3 )*( x - 9 ) = 0 }}}
{{{ x = 3 }}}
{{{ x = 9 }}}
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Only {{{ x=9 }}} is valid. If you plug in
{{{ x = 3 }}}, then
{{{  log(9,(3-6)) + log(9,(3-6)) = 1 }}}
{{{  log(9, (-3)) + log(9, (-3)) = 1 }}}
You can't have a log to a positive base
that gives you a negative number