Question 668792
I'll do #1 to get you started.


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


ln( x(x + 3) ) = 1


x(x + 3) = e^1


x(x + 3) = e


x^2 + 3x = e


x^2 + 3x - e = 0


x^2 + 3x - 2.71828 = 0   (note: e is roughly 2.71828...)


Now use the quadratic formula to solve for x. In this case, a = 1, b = 3, and c = 2.71828


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


{{{x = (-(3)+-sqrt((3)^2-4(1)(-2.71828)))/(2(1))}}}


{{{x = (-3+-sqrt(9-(-10.87312)))/(2)}}}


{{{x = (-3+-sqrt(9+10.87312))/(2)}}}


{{{x = (-3+-sqrt(19.87312))/2}}}


{{{x = (-3+sqrt(19.87312))/2}}} or {{{x = (-3-sqrt(19.87312))/2}}}


{{{x = (-3+4.4579278)/2}}} or {{{x = (-3-4.4579278)/2}}}


{{{x = 0.7289639}}} or {{{x = -3.7289639}}}


So the possible answers are {{{x = 0.7289639}}} or {{{x = -3.7289639}}}


However, plugging in {{{x = -3.7289639}}} yields a negative argument in the log, which is not allowed. So we must toss out this x value.


So the only solution (to 3 decimal places) is {{{x = 0.729}}}