Question

Log1/5 (x + 2) > log1/5 (- x)

Answer 1

Ответ:

решение смотри на фотографии

Answer 2

To solve the inequality $\log_{\frac{1}{5}}(x + 2) > \log_{\frac{1}{5}}(-x)$, we can start by simplifying the inequality using logarithm properties.

First, recall the logarithm property $\log_b(a) > \log_b(c)$ is equivalent to $a > c$ where $b > 1$. In this case, the base of the logarithm is $\frac{1}{5}$, which is between 0 and 1, so we need to reverse the inequality sign when we simplify the inequality.

Using this property, the inequality $\log_{\frac{1}{5}}(x + 2) > \log_{\frac{1}{5}}(-x)$ simplifies to:

$x + 2 > -x$

Now, let's solve for $x$:

$x + x > -2$ $2x > -2$

Divide by 2:

$x > -1$

So the solution to the inequality $\log_{\frac{1}{5}}(x + 2) > \log_{\frac{1}{5}}(-x)$ is $x > -1$.