Question 4558
{{{ln (x+1) = ln (3x + 1) - ln (x)}}}


By the second law of logarithms, {{{ln M - ln N = ln (M/N)}}}
{{{ln (x+1) = ln ((3x + 1)/x) }}}


Now, since there are ln s on both sides of the equation, raise both sides as a power of e:

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


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


Multiply both sides of this equation by x

{{{x(x+1) = 3x + 1}}}
{{{x^2 + x = 3x + 1}}}


Set equal to zero:
{{{x^2 - 2x - 1 = 0 }}}


Solve by quadratic formula, where {{{x = (-b+- sqrt (b^2 - 4ac))/(2a)}}} and a= 1, b= -2, c = -1.


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

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

{{{x = (2 +- sqrt (8) )/ 2 }}}

{{{x = (2 +- 2* sqrt (2)) / 2 }}}


Factor the numerator, so you can reduce the fraction:
{{{x = (2(1 +-sqrt (2)))/2 }}}

{{{x =1 + sqrt (2)  }}} or {{{x= 1 - sqrt (2)}}}


However, the second solution is a negative number, which is NOT allowed in the last ln expression.  So the only solution is {{{x = 1 + sqrt (2)}}}


R^2 at SCC