Question 627187
You must convert the right side into
a log expression.
Note that {{{ log(100) = 2 }}}, so
{{{ log(2y) + Log(4y-1) = 2 }}} 
{{{ log(2y) + Log(4y-1) = log(100) }}}
Since the general rule is: {{{ log( a*b ) = log(a) + log(b) }}}
{{{ log( (2y)*(4y-1) ) = log(100) }}}
{{{ (2y)*(4y-1) = 100 }}}
{{{ 8y^2 - 2y = 100 }}}
{{{ 8y^2 - 2y - 100 = 0 }}}
{{{ 4y^2 - y - 50 = 0 }}}
Use quadratic formula
{{{ y = (-b +- sqrt( b^2 - 4*a*c ))/(2*a) }}}
{{{ a = 4 }}}
{{{ b = -1 }}}
{{{ c = -50 }}}
{{{ y = (-(-1) +- sqrt( (-1)^2 - 4*4*(-50) ))/(2*4) }}}
{{{ y = ( 1 +- sqrt( 1 + 800 )) / 8 }}} 
{{{ y = ( 1 +- 28.302 ) / 8 }}}
{{{ y = 29.302/8 }}}
{{{ y = 3.663 }}}
and
{{{ y = -27.302/8 }}}
{{{ y = -3.413 }}}
check answers:
{{{ log(2y) + Log(4y-1) = 2 }}} 
Right away, I see the minus answer can't 
work, since you can't have a log base 10
that gives a negative number
{{{ log( 2*3.663 ) + log( 4*3.663 - 1 ) = 2 }}}
{{{ log( 7.326 ) + log( 13.652 ) = 2 }}}
{{{ .8649 + 1.1352  = 2 }}}
{{{ 2.000096 = 2 }}}
OK -error due to rounding off