Question 759703
Make the conversion:
{{{ 1 = log( 3, 3 ) }}}
{{{ log( 3, 2x - 1 ) + log( 3, x - 1 ) < log( 3, 3 ) }}}
{{{ log( 3, ( 2x - 1 )*( x - 1 )) < log( 3,3 ) }}}
{{{ ( 2x - 1 )*( x - 1 ) < 3 }}}
{{{ 2x^2 - x - 2x + 1  < 3 }}}
{{{ 2x^2 - 3x - 2 < 0 }}}
Use quadratic formula
{{{ x = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = 2 }}}
{{{ b = -3 }}}
{{{ c = -2 }}}
{{{ x = (-(-3) +- sqrt( (-3)^2 - 4*2*(-2) )) / (2*2) }}} 
{{{ x = ( 3 +- sqrt( 9 + 16 )) / 4 }}} 
{{{ x = ( 3 +- sqrt( 25 )) / 4 }}} 
{{{ x = ( 3 + 5) / 4 }}} 
{{{ x = 2 }}}
and
{{{ x = ( 3 - 5 ) / 4 }}}
{{{ x = -1/2 }}}
---------------
{{{ x - 2 < 0 }}}
{{{ x < 2 }}}
and
{{{ x + 1/2 < 0 }}}
{{{ x < -1/2 }}}
-------------
{{{ x < 2 }}} is the answer since it covers both cases
Hope I got it