Question 627462
{{{ln(x+3)+ ln(x)=1}}}


{{{ln(x(x+3))=1}}}


{{{x(x+3)=e^1}}}


{{{x(x+3)=e}}}


{{{x(x+3) = e}}}


{{{x^2 + 3x = e}}}


{{{x^2 + 3x - e = 0}}}


Now use the quadratic formula to solve for x


{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(3)+-sqrt((3)^2-4(1)(-e)))/(2(1))}}}


{{{x = (-3+-sqrt(9+4e))/(2)}}}


{{{x = (-3+sqrt(9+4e))/(2)}}} or {{{x = (-3-sqrt(9+4e))/(2)}}}


So the possible solutions are {{{x = (-3+sqrt(9+4e))/(2)}}} or {{{x = 
(-3-sqrt(9+4e))/(2)}}}


which approximate to x = 0.72896 or x = -3.72896


However, notice how the second possible solution is negative...which is not 
allowed.


So we toss that solution.


Therefore, the only solution is approximately x = 0.72896