Question 697789
The trick here is to express {{{ 2 }}} as a log
to the base {{{ 6 }}}:
{{{ 2 = log(6,36) }}}
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given:
{{{ log( 6,( x+5 ) ) + log( 6,( x + 4 ) ) = log( 6, 36 ) }}}
{{{ log( 6, ( x+5 )*( x+4 ) ) = log( 6, 36 ) }}}
{{{ ( x+5 )*( x+4 ) = 36 }}}
{{{ x^2 + 9x + 20 = 36 }}}
{{{ x^2 + 9x - 16 = 0 }}}
Use quadratic formula
{{{ x = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = 1 }}}
{{{ b = 9 }}}
{{{ c = -16 }}}
{{{ x = ( -9 +- sqrt( 9^2 - 4*1*(-16) )) / (2*1) }}} 
{{{ x = ( -9 +- sqrt( 81 + 64 )) / 2 }}} 
{{{ x = ( -9 +- sqrt( 145 )) / 2 }}} 
{{{ x = ( -9 + 12.042 ) / 2 }}}
{{{ x = 3.042 / 2 }}}
{{{ x = 1.521 }}}
I don't think you can use the negative square root of {{{ 145 }}}
at all because you end up with logs of negative numbers, 
and that is impossible.
check answer:
{{{ log( 6,( 1.521+5 ) ) + log( 6,( 1.521 + 4 ) ) = 2 }}}
{{{ log( 6, 6.521 ) + log( 6, 5.521 ) = 2 }}}
{{{ 1.048 + .96 = 2 }}}
Pretty close