Search
Ask a question
0

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

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

1 Answer by Expert Tutors

Tutors, sign in to answer this question.
Mark M. | Math Tutor--High School/College levelsMath Tutor--High School/College levels
4.9 4.9 (819 lesson ratings) (819)
0
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,∞)