To solve the inequality \( \log_{\frac{1}{3}}(x^2 - 3x + 5) < -1 \), we will follow these steps:
### Step 1: Understand the logarithmic inequality
The base of the logarithm is \( \frac{1}{3} \), which is between 0 and 1. Therefore, the logarithmic function is decreasing. This means that when we take the antilogarithm, the direction of the inequality will change.
### Step 2: Rewrite the inequality
We can rewrite the inequality as:
\[
x^2 - 3x + 5 > \left(\frac{1}{3}\right)^{-1}
\]
Calculating \( \left(\frac{1}{3}\right)^{-1} \) gives us \( 3 \). Thus, the inequality becomes:
\[
x^2 - 3x + 5 > 3
\]
### Step 3: Simplify the inequality
Subtract 3 from both sides:
\[
x^2 - 3x + 5 - 3 > 0
\]
This simplifies to:
\[
x^2 - 3x + 2 > 0
\]
### Step 4: Factor the quadratic expression
Next, we factor the quadratic expression:
\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\]
So, we have:
\[
(x - 1)(x - 2) > 0
\]
### Step 5: Determine the critical points
The critical points from the factors are \( x = 1 \) and \( x = 2 \). We will analyze the sign of the product \( (x - 1)(x - 2) \) in the intervals determined by these points: \( (-\infty, 1) \), \( (1, 2) \), and \( (2, \infty) \).
### Step 6: Test the intervals
1. **Interval \( (-\infty, 1) \)**: Choose \( x = 0 \):
\[
(0 - 1)(0 - 2) = (-1)(-2) = 2 > 0
\]
So, this interval satisfies the inequality.
2. **Interval \( (1, 2) \)**: Choose \( x = 1.5 \):
\[
(1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 < 0
\]
So, this interval does not satisfy the inequality.
3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \):
\[
(3 - 1)(3 - 2) = (2)(1) = 2 > 0
\]
So, this interval satisfies the inequality.
### Step 7: Combine the results
The solution to the inequality \( (x - 1)(x - 2) > 0 \) is:
\[
x \in (-\infty, 1) \cup (2, \infty)
\]
### Final Answer
Thus, the solution to the inequality \( \log_{\frac{1}{3}}(x^2 - 3x + 5) < -1 \) is:
\[
x \in (-\infty, 1) \cup (2, \infty)
\]
