Question 385066
Solve: log[2](3x+89)-log[2](x+8)=log[2](x+3)

     log[2](3x+89)-log[2](x+8)=log[2](x+3)
    =log[2](3x+89)-log[2](x+8)-log[2](x+3)=0
    =log[2](3x+89)-(log[2](x+8)+log[2](x+3)=0
    =log[2](3x+89)-(log[2](x+8)*(x+3)=0
     log[2](3x+89)=(log[2](x+8)*(x+3)
     3x+89=(x+8)*(x+3)
     3x+89=x^2+11x+24
       x^2+8x-65

   Use following  quadratic formula to solve with a=1, b=8, c=-65

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

     x = (-8ħsqrt(8^2-4*1*-65))/2*1
       = (-8ħsqrt(324)/2) 
       = (-8ħ13)/2 
     x = 5/2 or -21/2(reject,log of a negative number not valid) 

     ans: x = 5/2 = 2.5