Question

Log 4 (2x+2)<log4(3x-2)

Answer 1

log₄(2x+2) < log₄(3x-2)
учтём, что данная логарифмическая функция возрастающая, учтём ОДЗ и составим систему неравенств:
2x + 2 > 0              x > -1           x > -1
3x - 2 > 0               x > 2/3         x > 2/3
2x + 2 < 3x -2, ⇒   -x < -4, ⇒     x > 4
Ответ: x > 4

Answer 2

To solve the inequality $\log_4(2x+2) < \log_4(3x-2)$, we need to find the range of values of $x$ that satisfy this inequality. The logarithmic inequality can be solved using the properties of logarithms.

The inequality can be rewritten as:

$\log_4(2x+2) - \log_4(3x-2) < 0$.

We can use the property of logarithms that states: $\log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right)$.

Applying this property, we get:

$\log_4\left(\frac{2x+2}{3x-2}\right) < 0$.

Now, let's find the critical points, where the expression inside the logarithm becomes zero or undefined:

1. $2x+2 = 0$: $2x = -2$, $x = -1$.

2. $3x-2 = 0$: $3x = 2$, $x = \frac{2}{3}$.

However, $\frac{2x+2}{3x-2}$ is undefined when $x = \frac{2}{3}$ since it results in division by zero. Therefore, $x = \frac{2}{3}$ is not a valid solution.

Now, we have two critical points: $x = -1$ and $x = \frac{2}{3}$. These critical points divide the number line into three intervals:

1. $x < -1$
2. $-1 < x < \frac{2}{3}$
3. $x > \frac{2}{3}$

Now, we need to test the inequality in each interval. We'll pick a test point from each interval and evaluate the expression $\frac{2x+2}{3x-2}$ to determine if it is positive or negative in that interval. We can use the fact that the logarithm of a positive number is positive, and the logarithm of a negative number is negative.

1. For $x < -1$: Let's pick $x = -2$ $\frac{2(-2)+2}{3(-2)-2} = \frac{-2}{-8} = \frac{1}{4} > 0$. So, $\log_4\left(\frac{2x+2}{3x-2}\right) > 0$ in this interval.

2. For $-1 < x < \frac{2}{3}$: Let's pick $x = 0$ $\frac{2(0)+2}{3(0)-2} = \frac{2}{-2} = -1 < 0$. So, $\log_4\left(\frac{2x+2}{3x-2}\right) < 0$ in this interval.

3. For $x > \frac{2}{3}$: Let's pick $x = 1$ $\frac{2(1)+2}{3(1)-2} = \frac{4}{1} = 4 > 0$. So, $\log_4\left(\frac{2x+2}{3x-2}\right) > 0$ in this interval.

Now we know the sign of the expression $\log_4\left(\frac{2x+2}{3x-2}\right)$