Question 691408
{{{ log(9,2x+1) - log(3,x-1) = 0}}}....convert to base {{{10}}}


 {{{log(2x+1)/ log(9) -log(x-1)/ log(3)= 0}}}


 {{{log((2x+1))/ log((3^2)) -log((x-1))/ log((3))= 0}}}


 {{{log((2x+1))/ 2*log((3)) -log((x-1))/ log((3))= 0}}}


 {{{(log((2x+1)) -2*log((x-1)))/ 2*log((3))= 0}}}


=> if {{{(log((2x+1)) -2*log((x-1)))=0}}} whole equation above is equal to zero


if {{{(log((2x+1)) -2*log((x-1)))=0}}}, than


 {{{log((2x+1)) =2*log((x-1))}}}


{{{log((2x+1)) =log(((x-1)^2))}}}


=> {{{(2x+1) =(x-1)^2}}}.....solve for {{{x}}}


{{{2x+1=x^2-2x+1}}}


{{{x^2-2x-2x-1+1=0}}}


{{{x^2-4x=0}}}


{{{x(x-4)=0}}}.......only if {{{(x-4)=0}}} whole equation  {{{x(x-4)=0}}}


so,  if {{{(x-4)=0}}} ...=> {{{x=4}}}


{{{ graph( 600, 600, -10, 10, -10, 10, log(9,2x+1) - log(3,x-1) ) }}}