Question 24553
 ln(x+1)=ln(3x+1)-ln(x)..
USING LN (A*B)=LN A + LN B ....AND
LN (A/B)=LN A - LN B......WE GET
LN (X+1)=LN {(3X+1)/X}
TAKING ANTI LOGS...
X+1=(3X+1)/X
X(X+1)=3X+1
X^2+X=3X+1
X^2+X-3X-1=0
X^2-2X-1=0
THE SOLUTION OF AX^2+BX+C=0 IS GIVEN BY
 {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}.HENCE
 {{{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 }}}
{{{x = (1 +sqrt(2)) }}}AS X CANNOT BE NEGATIVE FOR LN X TO BE REAL