Question 691417
{{{ log(x+3)-log(x-3)=log(x-1) }}}
{{{ log((x+3)/(x-3))=log(x-1) }}}         (using {{{ log(a)-log(b) = log(a/b) }}})
{{{ (x+3)/(x-3)=x-1 }}}         (taking anti-logarithms)
{{{ x+3 = (x-3)(x-1) }}} 
{{{ x+3 = x^2-4x+3 }}} 
{{{ x^2-4x+3 = x+3 }}} 
{{{ x^2-5x=0 }}} 
{{{ x(x-5)=0 }}} 
so {{{ x=0 }}} or {{{ x=5 }}}

But original question had {{{ log(x-3) }}} so {{{ x-3 >0 }}} or  {{{ x>3 }}}

so, solution is  {{{ x=5 }}}