1+Log2 (3x+1) = Log2 (x^2-5)

1. ОДЗ:

{3x + 1 > 0;

{x² - 5 > 0; {x > - 1/3;

{ (x + √5) (x - √5) > 0; {x ∈ (-1/3; ∞);

{x ∈ (-∞; - √5) ∪ (√5; ∞). x ∈ (√5; ∞).

2. Решим:

1 + log2 (3x + 1) = log2 (x² - 5); log2 (2) + log2 (3x + 1) = log2 (x² - 5); log2 (2 (3x + 1)) = log2 (x² - 5); 2 (3x + 1) = x² - 5; 6x + 2 - x² + 5 = 0; 6x - x² + 7 = 0; x² - 6x - 7 = 0.

3. Четверть дискриминанта:

D/4 = (b/2) ² - ac; D/4 = (-3) ² - 1 * (-7) = 9 + 7 = 16; x = (-b/2 ± √ (D/4)) / a; x = (3 ± √16) / 1 = 3 ± 4; x1 = 3 - 4 = - 1 ∉ (√5; ∞); x2 = 3 + 4 = 7 ∈ (√5; ∞).

Ответ: 7.