Question

2log2(-x)=1+log 2(x+4)

Answer 1

Ответ на рисунке ниже.

Answer 2

To solve the equation 2log2(-x) = 1 + log2(x + 4), we need to find the value(s) of x that satisfy the equation. However, there's an issue with the equation as it stands. The domain of the logarithmic function is restricted to positive numbers. In this case, log2(-x) is not defined for any real number x, as it requires a positive base. Therefore, the equation has no real solutions.

Let's analyze the equation step-by-step to see why:

1. 2log2(-x) = 1 + log2(x + 4)

2. Rewrite the equation using properties of logarithms: log2((-x)^2) = log2(2) + log2(x + 4)

3. Simplify the equation further: log2(x^2) = log2(2) + log2(x + 4)

4. Since log2(x^2) and log2(2) are both constants, we can combine them: log2(x^2) = log2(2 * (x + 4))

5. Now, if we try to remove the logarithm, we get: x^2 = 2 * (x + 4)

6. Expanding the right side: x^2 = 2x + 8

7. Rearrange the equation to set it to zero: x^2 - 2x - 8 = 0

To solve this quadratic equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

where a = 1, b = -2, and c = -8.

Calculating the discriminant (b^2 - 4ac): Discriminant = (-2)^2 - 4 * 1 * (-8) = 4 + 32 = 36

Since the discriminant is positive, there are two distinct real solutions for x: x = (2 + √36) / 2 or x = (2 - √36) / 2 x = (2 + 6) / 2 or x = (2 - 6) / 2 x = 8 / 2 or x = -4 / 2 x = 4 or x = -2

However, upon evaluating these solutions in the original equation, we encounter an issue with the domain restriction for logarithms. Since log2(-x) is not defined for any real number x, there are no valid solutions for the given equation in the real number domain.