Question 389384
 <pre><font size = 3 color = "indigo"><b>
Hi
2log(base 2) x=3+log(base 2) (x+16
*[tex \LARGE\ \ 2log_2x - log_2(x+16) = log_2 x^2/(x+16)= 3 ]
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
 2^3 = x^2/(x+16)
 8(x+16) = x^2
 x^2 -8x - 128 = 0
either by factoring or using the Quadratic formula {{{x = (8 +-sqrt(576))/2}}}
  (x+8) = 0
  x = -8  this is an extraneous solution
  (x-16) = 0
  x = 16