Question 469443
{{{log((2x))+ log((2x-1))=1}}}
{{{system(2x>0,2x-1>0)}}}=> {{{system(x>0,2x>1)}}}=>{{{system(x>0,x>1/2)}}}=>{{{x>1/2}}}
Use formula {{{log(c,a)+log(c,b)=log(c,(a*b))}}}
{{{log(2x*(2x-1)) =1}}}
Use formula {{{log(c,a)=b}}}<=>{{{a=c^b}}}
{{{2x*(2x-1) =10^1}}}
{{{2x*(2x-1) =10}}}divide by 2
{{{x(2x-1)=5}}}
{{{2x^2-x-5=0}}}
 {{{x = (-(-1) +- sqrt( (-1)^2-4*2*(-5) ))/(2*2) }}} 
 {{{x = (1 +- sqrt( 41))/4 }}} 
{{{x = (1 - sqrt( 41))/4 =-1.35<1/2}}} extraneous root
{{{x = (1 + sqrt( 41))/4=1.85 }}}