Solve the following inequation .
(ii) `log_(2x)(x^2-5x+6)lt1`

Answer

To solve the inequation \( \log_{2x}(x^2 - 5x + 6) < 1 \), we will follow a systematic approach.

### Step 1: Understand the Logarithmic Inequality
The inequality \( \log_{2x}(x^2 - 5x + 6) < 1 \) implies that:
\[ x^2 - 5x + 6 < (2x)^1 \]
This is because \( \log_a(b) < c \) can be rewritten as \( b < a^c \) when \( a > 1 \).

### Step 2: Find Conditions for the Logarithm
Since the logarithm is defined only for positive arguments, we need:
\[ x^2 - 5x + 6 > 0 \]
To solve this quadratic inequality, we can factor it:
\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]
We need to find where this expression is greater than 0. The roots are \( x = 2 \) and \( x = 3 \).

### Step 3: Analyze the Sign of the Quadratic
Using a number line, we can test intervals:
- For \( x < 2 \) (e.g., \( x = 1 \)): \( (1 - 2)(1 - 3) = (-)(-) = + \) (positive)
- For \( 2 < x < 3 \) (e.g., \( x = 2.5 \)): \( (2.5 - 2)(2.5 - 3) = (+)(-) = - \) (negative)
- For \( x > 3 \) (e.g., \( x = 4 \)): \( (4 - 2)(4 - 3) = (+)(+) = + \) (positive)

Thus, the solution to \( x^2 - 5x + 6 > 0 \) is:
\[ x \in (-\infty, 2) \cup (3, \infty) \]

### Step 4: Consider the Base of the Logarithm
Next, we need to ensure that the base \( 2x > 0 \) and \( 2x \neq 1 \):
1. \( 2x > 0 \) implies \( x > 0 \).
2. \( 2x = 1 \) implies \( x = \frac{1}{2} \).

### Step 5: Solve the Inequality
Now we rewrite the inequality:
\[ x^2 - 5x + 6 < 2x \]
Rearranging gives:
\[ x^2 - 7x + 6 < 0 \]
Factoring yields:
\[ (x - 1)(x - 6) < 0 \]
Using a number line again:
- For \( x < 1 \) (e.g., \( x = 0 \)): \( (0 - 1)(0 - 6) = (-)(-) = + \) (positive)
- For \( 1 < x < 6 \) (e.g., \( x = 2 \)): \( (2 - 1)(2 - 6) = (+)(-) = - \) (negative)
- For \( x > 6 \) (e.g., \( x = 7 \)): \( (7 - 1)(7 - 6) = (+)(+) = + \) (positive)

Thus, the solution to \( (x - 1)(x - 6) < 0 \) is:
\[ x \in (1, 6) \]

### Step 6: Combine the Conditions
Now we combine the intervals from our findings:
1. From \( x^2 - 5x + 6 > 0 \): \( x \in (-\infty, 2) \cup (3, \infty) \)
2. From \( (x - 1)(x - 6) < 0 \): \( x \in (1, 6) \)

The intersection of these intervals is:
- From \( (-\infty, 2) \): \( (1, 2) \)
- From \( (3, \infty) \): \( (3, 6) \)

Thus, the final solution is:
\[ x \in (1, 2) \cup (3, 6) \]