Question

X^2*log625 (x-6) <= log5 (x^2-12x+36)

Answer 1

Ответ:

Пошаговое объяснение:

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Answer 2

To solve the inequality $x^2 \cdot \log_{625}(x-6) \leq \log_5(x^2 - 12x + 36)$, we need to follow a step-by-step process.

First, let's simplify the expressions using logarithmic properties:

1. Use the change of base formula to rewrite the logarithms: $\log_{625}(x-6) = \frac{\log_5(x-6)}{\log_5(625)}$ and $\log_5(x^2 - 12x + 36)$.

2. Simplify the expressions further: $\log_{625}(x-6) = \frac{1}{4} \log_5(x-6)$ (since $\log_5(625) = 4$).

Now, the inequality becomes: $x^2 \cdot \frac{1}{4} \log_5(x-6) \leq \log_5(x^2 - 12x + 36)$.

Multiply both sides of the inequality by 4 to eliminate the fraction: $x^2 \cdot \log_5(x-6) \leq 4 \cdot \log_5(x^2 - 12x + 36)$.

Now, we have: $x^2 \cdot \log_5(x-6) \leq \log_5((x-6)^4)$.

Using the property $\log_a(b^c) = c \cdot \log_a(b)$: $x^2 \cdot \log_5(x-6) \leq 4 \cdot \log_5(x-6)$.

Divide both sides of the inequality by $\log_5(x-6)$, noting that $\log_5(x-6)$ must be positive since the logarithm of a negative number is undefined: $x^2 \leq 4$.

Take the square root of both sides (keeping in mind that we need to consider both the positive and negative square roots): $x \leq 2$ or $x \geq -2$.

However, we need to make sure that the logarithms in the original expressions are well-defined. The logarithm $\log_5(x-6)$ must be positive, so $x > 6$. Additionally, in the expression $\log_5(x^2 - 12x + 36)$, we need to make sure the argument inside the logarithm is positive, so $x^2 - 12x + 36 > 0$, which simplifies to $(x - 6)^2 > 0$, which is always true for real $x$ except at $x = 6$.

So, considering all conditions together, the solution to the inequality is: $x \in (-\infty, -2] \cup (6, 2].$