Question 78142
You cannot foil these parenthesis, you must use logarithmic identities to solve

{{{lnx + ln(x+1)=2}}}

{{{ln(x*(x+1))=2}}} Use the identity: {{{ln(x)+ln(y)=ln(x*y)}}}

{{{ln(x^2+x)=2}}} Distribute

{{{e^(ln(x^2+x))=e^2}}} Raise both sides to exponents with the base of e. This undoes the natural log

{{{x^2+x=e^2}}}

{{{x^2+x=e^2}}} Subtract {{{e^2}}} from both sides

{{{x^2+x-e^2=0}}} Since e^2 is approximately 7.3891 we can say

{{{x^2+x-7.3891=0}}}

Now use the quadratic formula:

*[invoke quadratic "x", 1, 1, -7.3891 ]


So our answer is 

{{{x=-3.26389218313595}}} or {{{x=2.26389218313595}}}