Question

Log2 x+log2 (x+6)=4

Answer 1

Ответ:

2

Пошаговое объяснение:

$log 2 (x) + log 2 (x+6) = 4\\ ODZ: \\ \left \{ {{x > 0} \atop {x > -6}} \right. \\ log 2 (x^{2} + 6x) = 4\\ x^{2} +6x - 16=0\\ x1=-8\\ x2=2\\ ODZ: x=2$

Answer 2

Ответ:

х=2.

Пошаговое объяснение:

Решение дано на фото.

Answer 3

To solve the equation log2(x) + log2(x + 6) = 4, we can use logarithmic properties to simplify it. Remember that the sum of logarithms with the same base is equivalent to the logarithm of the product. Therefore, we can rewrite the equation as a single logarithm:

log2(x(x + 6)) = 4.

Now, we can remove the logarithm by converting the equation to exponential form:

2^4 = x(x + 6).

Simplifying further:

16 = x^2 + 6x.

Rearranging the equation to the quadratic form:

x^2 + 6x - 16 = 0.

Now, we can solve this quadratic equation. We can either factorize it, complete the square, or use the quadratic formula. In this case, let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a),

where a = 1, b = 6, and c = -16. Plugging in these values:

x = (-6 ± √(6^2 - 4(1)(-16))) / (2(1)).

Simplifying:

x = (-6 ± √(36 + 64)) / 2, x = (-6 ± √100) / 2, x = (-6 ± 10) / 2.

We have two possible solutions:

1. x = (-6 + 10) / 2 = 2.
2. x = (-6 - 10) / 2 = -8.

Therefore, the solutions to the equation log2(x) + log2(x + 6) = 4 are x = 2 and x = -8.