Question 729721


if {{{log(3, (3x+18))=log(3, (x^2+x))}}}

than

{{{3x+18=x^2+x}}}....solve for {{{x}}}

{{{0=x^2-3x+x-18}}}

{{{x^2-2x-18=0}}}.....use quadratic formula


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}...note, {{{a=1}}}, {{{b=-2}}}, and {{{c=-18}}}


{{{x = (-(-2) +- sqrt( (-2)^2-4*1*(-18) ))/(2*1) }}}



{{{x = (2 +- sqrt( 4+72 ))/2 }}}


{{{x = (2 +- sqrt( 76 ))/2 }}}


{{{x = (2 +- sqrt(4*19 ))/2 }}}


{{{x = (2 +- 2sqrt(19 ))/2 }}}

{{{x = (cross(2)1 +- cross(2)sqrt(19 ))/cross(2) }}}


{{{x = (1 +- sqrt(19 )) }}}

{{{x = (1 +- 4.35889894354) }}}

solutions:

{{{x = 1 + 4.35889894354 }}}

{{{x =5.35889894354 }}}


{{{x =5.36 }}}

or

{{{x = 1 - 4.35889894354}}}

{{{x = -3.35889894354}}}

{{{x = -3.36}}}