Question 617482
{{{ log(11,x-6) + log(11,x-4) = 1 }}}
Use the general rule:
{{{ log(a) + log(b) = log(a*b) }}}
{{{ log(11,x-6) + log(11,x-4) = log(11,(x-6)*(x-4)) }}}
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In the inverse notation, what this tells me is:
{{{ 11^1 = (x-6)*(x-4) }}}
{{{ x^2 - 10x + 24 = 11 }}}
{{{ x^2 - 10x + 13 = 0 }}}
Use the quadratic formula to solve:
{{{ x = ( -b +- sqrt( b^2-4*a*c )) / (2*a) }}} 
{{{ a = 1 }}}
{{{ b = -10 }}}
{{{ c = 13 }}}
{{{ x = ( -(-10) +- sqrt( (-10)^2 - 4*1*13 )) / (2*1) }}} 
{{{ x = ( 10 +- sqrt( 100 - 52 )) / 2 }}} 
{{{ x = ( 10 +- sqrt( 48 )) / 2 }}} 
{{{ x =  5 +- 2*sqrt(3) }}}
check:
{{{ log(11,5 + 2*sqrt(3) - 6) + log(11,5 + 2*sqrt(3) - 4) = 1 }}}
{{{ log(11,2*sqrt(3) - 1) + log(11,2sqrt(3) + 1) = 1 }}}
{{{ log(11, (2*sqrt(3) - 1)*(2*sqrt(3) + 1)) = 1 }}}
{{{ log(11,12 - 1) = 1 }}}
{{{ log(11,11) = 1 }}}