log(2;x) + log(2;(x+1))>=0

Assuming log base 2, i.e., log2 x.

• Domain: x > 0 and x+1 > 0 → x > 0.
• Use log(a) + log(b) = log(ab): log2 x + log2 (x+1) = log2[x(x+1)].
• Since base 2 > 1, log2[y] ≥ 0 ⇔ y ≥ 1 (with y > 0).
• So require x(x+1) ≥ 1 → x^2 + x − 1 ≥ 0.
• Solve: x^2 + x − 1 = 0 has roots x = [−1 ± √5]/2 = −(1+√5)/2 and (√5−1)/2. The inequality holds for x ≤ −(1+√5)/2 or x ≥ (√5−1)/2.
• Intersect with domain x > 0 → x ∈ [ (√5 − 1)/2, ∞ ).

Numerical value: (√5 − 1)/2 ≈ 0.618.

Answer: x ≥ (√5 − 1)/2.