Question

Log^3 (x^2 - 2x + 1) меньше или равно 2

Answer 1

У меня такое же Д/З только мне задали 1 и 2

Answer 2

To solve the inequality $\log^3(x^2 - 2x + 1) \leq 2$, we need to find the values of $x$ that satisfy the inequality. Since the base of the logarithm is not specified, we assume that it is the common logarithm with base 10.

Here are the steps to solve the inequality:

Step 1: Rewrite the inequality using the properties of logarithms. $\log^3(x^2 - 2x + 1) \leq 2$

Step 2: Convert the inequality to an exponential form. Remember that if $\log_b(y) = z$, then $b^z = y$. $10^2 \leq x^2 - 2x + 1$

Step 3: Simplify the exponential expression on the right side. $100 \leq x^2 - 2x + 1$

Step 4: Move all terms to one side of the inequality to set it to zero. $x^2 - 2x + 1 - 100 \leq 0$

Step 5: Combine like terms and factorize if possible. $x^2 - 2x - 99 \leq 0$

Step 6: Solve the quadratic inequality. To do this, we can find the critical points by setting the expression inside the inequality to zero: $x^2 - 2x - 99 = 0$ We can factor this quadratic equation: $(x - 11)(x + 9) = 0$ This gives us two critical points: $x = 11$ and $x = -9$.

Step 7: Analyze the inequality in intervals using a sign chart or test points.

• Interval 1: $x < -9$ Let's take a test point $x = -10$ (any value less than -9) and plug it into the inequality: $(-10)^2 - 2(-10) - 99 \leq 0$ $100 + 20 - 99 \leq 0$ $21 \leq 0$ (False)

• Interval 2: $-9 < x < 11$ Let's take a test point $x = 0$ (any value between -9 and 11) and plug it into the inequality: $0^2 - 2(0) - 99 \leq 0$ $-99 \leq 0$ (True)

• Interval 3: $x > 11$ Let's take a test point $x = 12$ (any value greater than 11) and plug it into the inequality: $12^2 - 2(12) - 99 \leq 0$ $144 - 24 - 99 \leq 0$ $21 \leq 0$ (False)

Step 8: Determine the solution set.

The solution to the inequality $x^2 - 2x - 99 \leq 0$ is $-9 \leq x \leq 11$. This means the inequality $\log^3(x^2 - 2x + 1) \leq 2$ is satisfied when $-9 \leq x \leq 11$.