Question 757101
{{{ log( 4, x-3 ) + log( 4, x+9 ) = 3 }}}
Make the substitution:
{{{ 3 = log( 4, 64 ) }}}
{{{ log( 4, x-3 ) + log( 4, x+9 ) = log( 4, 64 ) }}}
{{{ log( 4,  ( x-3 )*( x+9 ) ) = log( 4, 64 ) }}}
{{{ ( x-3 )*( x+9 ) = 64 }}}
{{{ x^2 - 3x + 9x - 27 = 64 }}}
{{{ x^2 + 6x - 91 = 0 }}}
Use quadratic formula
{{{ x = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = 6 }}}
{{{ c = -91 }}}
{{{ x = (-6 +- sqrt( 6^2 - 4*1*(-91) )) / (2*1) }}}
{{{ x = (-6 +- sqrt( 36 + 364 )) / 2 }}}
{{{ x = (-6 +- sqrt( 400 )) / 2 }}}
{{{ x = ( -6 + 20 ) / 2 }}}
{{{ x = 7 }}}
and
{{{ x = ( -6 - 20 ) / 2 }}}
{{{ x = -13 }}}
This negative answer must be rejected because a
log to the base {{{ 4 }}} cannot give you a negative result
{{{ x = 7 }}} is the answer
check:
{{{ log( 4, x-3 ) + log( 4, x+9 ) = 3 }}}
{{{ log( 4, 7-3 ) + log( 4, 7+9 ) = 3 }}}
{{{ log( 4, 4 ) + log( 4, 16 ) = 3 }}}
{{{ 1 + 2 = 3 }}}
OK