Question 136175
{{{log(6,(x+1))+log(6,(x))=1}}}
{{{log(6,(x(x+1)))=1}}}
{{{6^(1)=x(x+1)}}}
{{{x^2+x=6}}}
{{{x^2+x-6=0}}}
{{{(x+3)(x-2)=0}}}
{{{x=2}}} and {{{x=-3}}}.
Check your answers.
{{{log(6,(x(x+1)))=1}}}
{{{log(6,(-3(-2)))=1}}}
{{{log(6,6)=1}}}
True statement, but only for the converted equation. 
In your original equation,
{{{log(6,(x+1))+log(6,(x))=1}}}
{{{log(6,(-2))+log(6,(-3))=1}}}
The solution x=-3 is not allowed, since the logarithm function must have positive arguments. 
{{{log(6,(2+1))+log(6,(2))=1}}}
{{{log(6,(3))+log(6,(2))=1}}}
{{{log(10,3)/log(10,6)+log(10,2)/log(10,6)=1}}}
{{{0.613147+0.386853=1}}}
{{{1=1}}}
True statement.
Good answer.
x=2.