Question 665932
{{{log((x))-log((x-1))=log((3x+12))}}}..use Quotient Rule 

{{{log((x))/log((x-1))=log((3x+12))}}}

{{{log((x/(x-1)))=log((3x+12))}}}


{{{log((x/(x-1)))=log((3x+12))}}}.....if log same, then


{{{(x/(x-1))=(3x+12)}}}.....solve for {{{x}}}


{{{x=(3x+12)(x-1)}}}


{{{x=3x^2-3x+12x-12}}}

{{{0=3x^2-3x-x+12x-12}}}

{{{3x^2+8x-12=0}}}....use quadratic formula and find positive root (it IS possible to calculate the log of a negative number but you have to work in complex numbers)

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-8 +- sqrt(8^2-4*3*(-12) ))/(2*3) }}}

{{{x = (-8 +- sqrt(64+144))/6 }}}

{{{x = (-8 +- sqrt(208))/6 }}}

{{{x = (-8 +- 14.42)/6 }}}

{{{x = 6.42/6 }}}

{{{x = 1.07 }}}


check:

{{{log(1.07)-log(1.07-1)=log(3*1.07+12)}}}

{{{log(1.07)-log(0.07)=log(3.21+12)}}}

{{{log(1.07)-log(0.07)=log(15.21)}}}

{{{0.029-(-1.154)=1.18}}}

{{{0.029+1.154=1.18}}}

{{{1.183=1.182}}}