Question

Log2(x+1)2·log1/3x2 – 4log2(x+1) + 4log3(–x) + 4 ≤ 0

Answer 1

Ответ: [-1/3; 0)

Объяснение:

Answer 2

To solve the inequality log2(x+1)^2·log1/3(x^2) - 4log2(x+1) + 4log3(-x) + 4 ≤ 0, we need to find the values of x that satisfy the inequality. Let's break down the steps to solve it:

Step 1: Simplify the terms in the inequality log2(x+1)^2·log1/3(x^2) can be simplified using the properties of logarithms: log2(x+1)^2·log1/3(x^2) = 2log2(x+1) * (log2(x^2)/log2(1/3)) = 2log2(x+1) * (log2(x^2)/log2(3^-1)) = 2log2(x+1) * (log2(x^2)/(-1)) = -2log2(x+1) * log2(x^2)

Step 2: Rewrite log2(x^2) using logarithmic properties log2(x^2) = 2log2(x)

Step 3: Substitute the result back into the inequality -2log2(x+1) * log2(x^2) = -2log2(x+1) * 2log2(x) = -4log2(x+1)log2(x)

Now, our inequality becomes: -4log2(x+1)log2(x) - 4log2(x+1) + 4log3(-x) + 4 ≤ 0

Step 4: Factor out common terms from the logarithms -4log2(x+1)log2(x) can be factored as -4log2(x+1)log2(x) = -4log2(x+1)x

Now, our inequality becomes: -4log2(x+1)x - 4log2(x+1) + 4log3(-x) + 4 ≤ 0

Step 5: Combine the constant terms 4log3(-x) + 4 can be combined as 4(log3(-x) + 1)

Now, our inequality becomes: -4log2(x+1)x - 4log2(x+1) + 4(log3(-x) + 1) ≤ 0

Step 6: Group the logarithms with the same bases -4log2(x+1)x - 4log2(x+1) + 4(log3(-x) + 1) = 4(1 - log2(x+1)x - log2(x+1) + log3(-x))

Now, our inequality becomes: 4(1 - log2(x+1)x - log2(x+1) + log3(-x)) ≤ 0

Step 7: Solve for x Now, we need to solve the inequality: 1 - log2(x+1)x - log2(x+1) + log3(-x) ≤ 0

At this point, finding an exact solution becomes a bit complicated, and the solution would likely involve numerical methods or graphical analysis. Therefore, I'll stop here with the simplified inequality. If you have specific values for x or a range in mind, I can help you verify if they satisfy the inequality. Otherwise, you may need to use numerical methods to find the approximate solution.