Question 703564
{{{ log( 2,8x ) - log( 2, (x^2 - 1) ) = log( 2,3 ) }}}
{{{ log( 2, (8x) / ( x^2 - 1 ) ) = log( 2,3 ) }}}
{{{ 8x / ( x^2 - 1 ) = 3 }}}
{{{ 3*( x^2 - 1 ) = 8x }}}
{{{ 3x^2 - 3 - 8x = 0 }}}
{{{ 3x^2 - 8x - 3 = 0 }}}
Use the quadratic formula
{{{ x = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 3 }}}
{{{ b = -8 }}}
{{{ c = -3 }}}
{{{ x = ( -(-8) +- sqrt( (-8)^2 - 4*3*(-3) )) / (2*3) }}}
{{{ x = ( 8 +- sqrt( 64 + 36 )) / 6 }}}
{{{ x = ( 8 +- sqrt( 100 )) / 6 }}}
{{{ x = ( 8 + 10 ) / 6 }}}
{{{ x = 18/6 }}}
{{{ x = 3 }}}
and
{{{ x = ( 8 - 10 ) / 6 }}}
{{{ x = -2/6 }}}
{{{ x = -1/3 }}}
This negative result can't be used because of
{{{ log( 2,8x ) }}}, which gives me
{{{ log(2, -8/3 ) }}}
The log to a positive base cannot give a negative result
The {{{ x = 3 }}} result is OK