Question 61746
ln(3x+7) + lnx = ln6
or ln((3x+7)*x) - ln6 = 0
or ln((3x+7)*x/6) = 0


This means,
{{{log(e,((3x+7)*x/6)) = 0}}}
or {{{(3x+7)*x/6 = e^0 = 1}}}
or {{{(3x+7)*x = 6}}}
or {{{3x^2 + 7x - 6 = 0}}}
or {{{3x^2 + 9x - 2x - 6 = 0}}}
or {{{3x(x + 3) - 2(x + 3) = 0}}}
or {{{(3x-2)(x + 3) = 0}}}


Either or {{{3x - 2 = 0}}} or {{{x + 3 = 0}}} 
i.e. either {{{x = 2/3}}} or {{{x = -3}}}


But, logarithm of a negative number is not defined. If {{{x = -3}}} then {{{log(e,x) = log(e,-3)}}} becomes undefined. So x cannot be -3.
Thus {{{x = 2/3}}} is the only solution.