Question

Log2(x+4)+log2(5-x)=3

Answer 1

Ответ:

Пошаговое объяснение:

log₂(x+4)+log₂(5-x)=3

сразу находим ООФ (x+4) >0 x > -4

(5-x) > 0 x < 5

x ∈ (-4; 5)

log₂((x+4)(5-x))=3

2³ = (x+4)(5-x)

5x-x²+20-4x=8

x² -x -12 = 0⇒ x₁ = 4; x₂= -3

оба корня лежат внутри промежутка (-4; 5)

x₁,x₂ ∈(-4; 5)

ответ

x₁ = 4; x₂= -3

Answer 2

To solve the equation log2(x + 4) + log2(5 - x) = 3, you can use logarithmic properties to simplify it and find the value of x. Here are the steps:

1. Combine the two logarithms on the left side using the product rule of logarithms:

   log2((x + 4)(5 - x)) = 3

2. Now, remove the logarithm by using the definition of logarithms. When the logarithm base is 2, you can rewrite the equation as:

   (x + 4)(5 - x) = 2^3

3. Simplify the equation:

   (x + 4)(5 - x) = 8

4. Expand the left side of the equation:

   5x - x^2 + 20 - 4x = 8

5. Combine like terms:

   5x - x^2 - 4x + 20 = 8

6. Simplify further:

   x^2 - 9x + 20 = 8

7. Subtract 8 from both sides:

   x^2 - 9x + 20 - 8 = 0

8. Combine constants:

   x^2 - 9x + 12 = 0

9. Factor the quadratic equation:

   (x - 3)(x - 4) = 0

10. Set each factor equal to zero and solve for x:

x - 3 = 0 or x - 4 = 0

x = 3 or x = 4

So, the solutions to the equation are x = 3 and x = 4. You should check both solutions to ensure they are valid by plugging them back into the original equation, as logarithms are only defined for positive values:

1. For x = 3: log2(3 + 4) + log2(5 - 3) = log2(7) + log2(2) = 3 + 1 = 4 (valid)

2. For x = 4: log2(4 + 4) + log2(5 - 4) = log2(8) + log2(1) = 3 + 0 = 3 (valid)

Both solutions are valid for the given equation.