Question

X^2log512(x+7)<=log2(x^2+14x+49)

Answer 1

Ответ:

(-бесконечность;-7] [ -6; + бесконечность)

Answer 2

To solve the inequality $x^2\log_{512}(x+7) \leq \log_2(x^2+14x+49)$, we can simplify the expression first. Since the logarithms have different bases, let's convert them to a common base, such as base 10.

Using the logarithmic property $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$, we have:

$\log_{512}(x+7) = \frac{\log_{10}(x+7)}{\log_{10}(512)}$ and $\log_2(x^2+14x+49) = \frac{\log_{10}(x^2+14x+49)}{\log_{10}(2)}$.

Substituting these values back into the inequality, we get:

$x^2\left(\frac{\log_{10}(x+7)}{\log_{10}(512)}\right) \leq \frac{\log_{10}(x^2+14x+49)}{\log_{10}(2)}$.

Simplifying further, we can cross-multiply to get rid of the denominators:

$x^2\log_{10}(x+7) \cdot \log_{10}(2) \leq \log_{10}(x^2+14x+49) \cdot \log_{10}(512)$.

Now, let's simplify the logarithms on both sides:

$x^2\log_{10}(x+7) \cdot \log_{10}(2) \leq \log_{10}(x^2+14x+49) \cdot \log_{10}(512)$.

Using the property $\log_a(b) \cdot \log_b(c) = \log_a(c)$, we can rewrite the inequality as:

$\log_{10}(x+7)^{x^2} \cdot \log_{10}(2)^{x^2} \leq \log_{10}(x^2+14x+49)^{512}$.

Now, let's raise both sides to the base 10 to simplify it further:

$(x+7)^{x^2} \cdot 2^{x^2} \leq (x^2+14x+49)^{512}$.

Solving this inequality analytically is quite complex due to the high exponent on the right side. It may not have a simple algebraic solution. To solve it numerically, you can use approximation methods or graph the functions to find the range of x values that satisfy the inequality.