Question

X^2log625(2-x)<=log5(x^2-4x+4)

Answer 1

Ответ:

Пошаговое объяснение:

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Answer 2

To solve the inequality $x^2 \log_{625}(2-x) \leq \log_5(x^2 - 4x + 4)$, we need to be careful with the logarithmic terms and the restrictions on the domain of the logarithmic functions. Let's break down the steps to solve this inequality:

Step 1: Domain Restrictions Notice that $\log_{625}(2-x)$ is only defined when $2 - x > 0$ because the base of the logarithm, 625, is a positive number. This means $x < 2$. Also, the expression $\log_5(x^2 - 4x + 4)$ is only defined when $x^2 - 4x + 4 > 0$, which simplifies to $(x-2)^2 > 0$. This is true for all real $x$ except $x = 2$.

So, the domain for this inequality is $x < 2$ and $x \neq 2$.

Step 2: Solve the Inequality Let's deal with the logarithmic terms first:

$x^2 \log_{625}(2-x) \leq \log_5(x^2 - 4x + 4)$

Since $x < 2$, $\log_{625}(2-x) > 0$ (since the base is greater than 1 and the argument is positive).

Now, we can rewrite the inequality as:

$x^2 \leq \frac{\log_5(x^2 - 4x + 4)}{\log_{625}(2-x)}$

But remember that $\log_{625}(2-x)$ is positive, so dividing by it doesn't change the inequality direction. Therefore, we have:

$x^2 \leq \frac{\log_5(x^2 - 4x + 4)}{\log_{625}(2-x)}$

Step 3: Solve the Numerator Let's simplify the right side of the inequality:

$\frac{\log_5(x^2 - 4x + 4)}{\log_{625}(2-x)} = \log_{625}(x^2 - 4x + 4)$

Now we have:

$x^2 \leq \log_{625}(x^2 - 4x + 4)$

Step 4: Evaluate the Logarithm Remember that $\log_{625}(x^2 - 4x + 4)$ is defined only when $(x-2)^2 > 0$, which is true for all $x$ except $x = 2$. But we have already established that $x = 2$ is not in the domain of the original inequality.

Therefore, the inequality holds true for all $x < 2$ (within the domain) because $x^2$ will always be non-negative and less than or equal to $\log_{625}(x^2 - 4x + 4)$ in that domain.

In summary, the solution to the inequality is $x < 2$, within the given domain restrictions.