Hi

2log(base 2) x=3+log(base 2) (x+16





 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) = 0

  x = -8  this is an extraneous solution

  (x-16) = 0

  x = 16

  

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