Question

Log2(x+1)+log2(x+2)=3-log2 (4)

Answer 1

ОДЗ: x + 1 > 0  ⇒ x > - 1
         x + 2 > 0  ⇒ x > - 2
Значит окончательно: x ∈ (- 1 ; + ∞)
$log _{2} (x+1)+log _{2} (x+2)=3-log _{2}4\\\\log _{2} (x+1)(x+2)=log_{2}8 -log _{2}4\\\\log _{2}(x+1)(x+2)=log _{2} \frac{8}{4} \\\\log _{2}(x+1)(x+2)=log _{2} 2\\\\(x+1)(x+2)=2\\\\ x^{2} +2x+x+2-2=0\\\\ x^{2} +3x=0\\\\ x(x+3)=0\\\\ x_{1} =0\\\\x+3=0\\\\ x_{2}=-3$
x₂ = - 3 - не подходит
Ответ : 0

Answer 2

To solve the equation:

log2(x + 1) + log2(x + 2) = 3 - log2(4)

We can simplify the right-hand side first. Recall that log2(4) = 2 because 2^2 = 4.

So the equation becomes:

log2(x + 1) + log2(x + 2) = 3 - 2

Now, combine the logarithms on the left-hand side using the logarithm product rule:

log2((x + 1)(x + 2)) = 1

Now we have a single logarithm on the left-hand side. To get rid of the logarithm, we can rewrite it as an exponential equation:

2^1 = (x + 1)(x + 2)

Simplify the right-hand side:

2 = x^2 + 3x + 2

Now bring everything to one side of the equation:

x^2 + 3x + 2 - 2 = 0

x^2 + 3x = 0

Now we have a quadratic equation. To solve for x, factorize it:

x(x + 3) = 0

This equation has two solutions:

1. x = 0
2. x = -3

However, we need to check if these solutions satisfy the original equation because logarithms can only be taken for positive numbers.

For x = 0:

log2(0 + 1) + log2(0 + 2) = log2(1) + log2(2) = 0 + 1 = 1

On the right-hand side:

3 - log2(4) = 3 - 2 = 1

The left-hand side and right-hand side match, so x = 0 is a valid solution.

For x = -3:

log2(-3 + 1) + log2(-3 + 2) = log2(-2) + log2(-1)

However, log2 of negative numbers is undefined in the real number domain, so x = -3 is not a valid solution.

Therefore, the only valid solution to the equation is x = 0.