Question 867256
a) {{{log((x+1)) = log(6)- log(x)}}}
{{{log((x+1)) = log(6/x)}}}
{{{x+1=6/x}}}
{{{x(x+1)=x(6/x)}}}
{{{x^2+x=6}}}
That is a quadratic equation.
It can be solved several ways, like factoring, using the quadratic formula or completing the square.
Completing the square:
{{{x^2+x=6}}}
{{{x^2+x+1/4=6+1/4}}}
{{{(x+1/2)^2=25/4}}} --> {{{system(x+1/2=5/2,"or",x+1/2=-5/2)}}}-->{{{system(x=5/2-1/2,"or",x=-5/2-1/2)}}}}-->{{{system(highlight(x=2),"or",x=-3)}}}
We discard {{{x=-3}}} because it makes {{{log(x)}}} undefined.
 
b) {{{ln(x-2) + ln(3) = ln(x+1)}}}
{{{ln((x-2)*3) = ln(x+1)}}}
{{{(x-2)*3 = x+1}}}
{{{3x-6 =x+1}}}
{{{3x-x =1+6)}}}
{{{2x=7}}}
{{{highlight(x=7/2)}}}