Question

Log2(3x+2)/log3(2x+3)<=0

Answer 1

Я точно не уверенна в этом, но думаю что это именно так решается. Скорее всего это так решается...Мне такие уравнения просто не попадались раньше 

Answer 2

To solve the inequality $\frac{\log_2(3x+2)}{\log_3(2x+3)} \leq 0$, we'll first find the critical points where the expression is equal to zero or undefined and then determine the sign of the expression in each interval between these critical points.

1. Find the critical points:

   • The numerator $\log_2(3x+2)$ is undefined for $3x + 2 \leq 0$, which means $x \leq -\frac{2}{3}$. So, we have a critical point at $x = -\frac{2}{3}$.

   • The denominator $\log_3(2x+3)$ is undefined for $2x + 3 \leq 0$, which means $x \leq -\frac{3}{2}$. So, we have another critical point at $x = -\frac{3}{2}$.

2. Create intervals based on the critical points:

   We now have three intervals to consider: $(-\infty, -\frac{3}{2})$, $(-{\frac{3}{2}}, -{\frac{2}{3}})$, and $(-{\frac{2}{3}}, \infty)$.

3. Analyze the sign of the expression in each interval:

   a. For $x < -\frac{3}{2}$: Both the numerator and denominator are negative. So, the expression is $\frac{\text{Negative}}{\text{Negative}}$, which is positive.

   b. For $-\frac{3}{2} < x < -\frac{2}{3}$: The numerator is negative (since $3x+2$ is positive), and the denominator is positive (since $2x+3$ is positive). So, the expression is $\frac{\text{Negative}}{\text{Positive}}$, which is negative.

   c. For $x > -\frac{2}{3}$: Both the numerator and denominator are positive. So, the expression is $\frac{\text{Positive}}{\text{Positive}}$, which is positive.

4. Combine the information:

   • The inequality is satisfied when the expression is less than or equal to zero. Therefore, the solution to the inequality is $x \in \boxed{\left(-\frac{3}{2}, -\frac{2}{3}\right]}$.

So, the solution to the inequality $\frac{\log_2(3x+2)}{\log_3(2x+3)} \leq 0$ is $x \in \left(-\frac{3}{2}, -\frac{2}{3}\right]$.