Solving Log2(2x)+log2(x+2)>log2(16) algebraically

Question

Solving Log2(2x)+log2(x+2)>log2(16) algebraically

Answer 1

log₂(2x) + log₂(x+2) > log₂16

 

  NOTE:  2x and x+2 must both be positive, so x > 0.

 

log₂[2x(x+2)] > log₂16

 

Since y = log₂x is an increasing function, we can conclude that 2x(x+2) > 16

 

     2x² + 4x - 16 > 0

 

       x² + 2x - 8 > 0

 

      (x+4)(x-2) > 0 and from the NOTE above, x > 0

 

      If x > 0 and x < 2, then (x+4)(x-2) < 0

      If x > 2, then (x+4)(x-2) > 0

 

Solution:  (2,∞)