To solve the inequality \( \log_2 \left( \frac{x-4}{2x+5} \right) < 1 \), we will follow these steps:
### Step 1: Determine the domain of the logarithmic function
The logarithm is defined when its argument is positive. Therefore, we need:
\[
\frac{x-4}{2x+5} > 0
\]
This implies that both the numerator \(x - 4\) and the denominator \(2x + 5\) must either be both positive or both negative.
### Step 2: Find the critical points
1. Set \(x - 4 = 0\) to find when the numerator is zero:
\[
x = 4
\]
2. Set \(2x + 5 = 0\) to find when the denominator is zero:
\[
2x + 5 = 0 \implies x = -\frac{5}{2}
\]
### Step 3: Test intervals around the critical points
We will test the intervals determined by the critical points \(x = -\frac{5}{2}\) and \(x = 4\):
- Interval 1: \( (-\infty, -\frac{5}{2}) \)
- Interval 2: \( (-\frac{5}{2}, 4) \)
- Interval 3: \( (4, \infty) \)
**Testing Interval 1: \(x = -3\)**
\[
\frac{-3 - 4}{2(-3) + 5} = \frac{-7}{-6 + 5} = \frac{-7}{-1} = 7 > 0 \quad \text{(Positive)}
\]
**Testing Interval 2: \(x = 0\)**
\[
\frac{0 - 4}{2(0) + 5} = \frac{-4}{5} < 0 \quad \text{(Negative)}
\]
**Testing Interval 3: \(x = 5\)**
\[
\frac{5 - 4}{2(5) + 5} = \frac{1}{10} > 0 \quad \text{(Positive)}
\]
### Step 4: Combine results from the intervals
From the tests, we find:
- The expression is positive in the intervals \( (-\infty, -\frac{5}{2}) \) and \( (4, \infty) \).
### Step 5: Solve the inequality \( \log_2 \left( \frac{x-4}{2x+5} \right) < 1 \)
This can be rewritten using the property of logarithms:
\[
\frac{x-4}{2x+5} < 2
\]
### Step 6: Rearranging the inequality
Rearranging gives:
\[
\frac{x-4}{2x+5} - 2 < 0 \implies \frac{x-4 - 2(2x + 5)}{2x + 5} < 0
\]
This simplifies to:
\[
\frac{x - 4 - 4x - 10}{2x + 5} < 0 \implies \frac{-3x - 14}{2x + 5} < 0
\]
### Step 7: Find critical points for the new inequality
Set the numerator and denominator to zero:
1. \( -3x - 14 = 0 \implies x = -\frac{14}{3} \)
2. \( 2x + 5 = 0 \implies x = -\frac{5}{2} \)
### Step 8: Test intervals for the new inequality
We will test the intervals determined by \(x = -\frac{14}{3}\) and \(x = -\frac{5}{2}\):
- Interval 1: \( (-\infty, -\frac{14}{3}) \)
- Interval 2: \( (-\frac{14}{3}, -\frac{5}{2}) \)
- Interval 3: \( (-\frac{5}{2}, \infty) \)
**Testing Interval 1: \(x = -5\)**
\[
\frac{-3(-5) - 14}{2(-5) + 5} = \frac{15 - 14}{-10 + 5} = \frac{1}{-5} < 0 \quad \text{(Negative)}
\]
**Testing Interval 2: \(x = -4\)**
\[
\frac{-3(-4) - 14}{2(-4) + 5} = \frac{12 - 14}{-8 + 5} = \frac{-2}{-3} > 0 \quad \text{(Positive)}
\]
**Testing Interval 3: \(x = 0\)**
\[
\frac{-3(0) - 14}{2(0) + 5} = \frac{-14}{5} < 0 \quad \text{(Negative)}
\]
### Step 9: Combine results from the new intervals
The expression is negative in the intervals \( (-\infty, -\frac{14}{3}) \) and \( (-\frac{5}{2}, \infty) \).
### Step 10: Combine both conditions
The solution must satisfy both conditions:
1. \( (-\infty, -\frac{5}{2}) \cup (4, \infty) \)
2. \( (-\infty, -\frac{14}{3}) \cup (-\frac{5}{2}, \infty) \)
The combined solution is:
\[
x \in (-\infty, -\frac{14}{3}) \cup (4, \infty)
\]
### Final Answer
\[
x \in (-\infty, -\frac{14}{3}) \cup (4, \infty)
\]
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