If `log_(0.3)(x-1) lt log_(0.09)(x-1)`, then x lies in the interval

Options

A. `(2,oo)`
B. `(-2,-1)`
C. `(1,2)`
D. none of these

Answer

To solve the inequality \( \log_{0.3}(x-1) < \log_{0.09}(x-1) \), we will follow these steps:

### Step 1: Rewrite the logarithm
We know that \( 0.09 \) can be expressed as \( 0.3^2 \). Therefore, we can rewrite the inequality as:
\[ \log_{0.3}(x-1) < \log_{0.3}( (x-1)^2 ) \]

### Step 2: Apply the properties of logarithms
Using the property of logarithms that states \( \log_b(a) < \log_b(c) \) if \( a < c \) when \( b < 1 \), we can rewrite the inequality:
\[ x-1 < (x-1)^2 \]

### Step 3: Rearrange the inequality
Rearranging the inequality gives us:
\[ 0 < (x-1)^2 - (x-1) \]
This simplifies to:
\[ 0 < (x-1)((x-1) - 1) = (x-1)(x-2) \]

### Step 4: Solve the inequality
Now, we need to determine when \( (x-1)(x-2) > 0 \). This product is positive when both factors are either positive or both are negative.

1. **Finding critical points**: The critical points are \( x = 1 \) and \( x = 2 \).
2. **Test intervals**: We will test the intervals determined by these critical points:
- Interval \( (-\infty, 1) \)
- Interval \( (1, 2) \)
- Interval \( (2, \infty) \)

### Step 5: Analyze the intervals
- For \( x < 1 \): Both \( (x-1) < 0 \) and \( (x-2) < 0 \), so \( (x-1)(x-2) > 0 \).
- For \( 1 < x < 2 \): \( (x-1) > 0 \) and \( (x-2) < 0 \), so \( (x-1)(x-2) < 0 \).
- For \( x > 2 \): Both \( (x-1) > 0 \) and \( (x-2) > 0 \), so \( (x-1)(x-2) > 0 \).

### Step 6: Combine the results
The inequality \( (x-1)(x-2) > 0 \) holds in the intervals:
- \( (-\infty, 1) \)
- \( (2, \infty) \)

### Step 7: Consider the domain of the logarithm
Since we have \( \log_{0.3}(x-1) \), we need \( x-1 > 0 \) or \( x > 1 \). Therefore, we discard the interval \( (-\infty, 1) \).

### Final Result
The solution to the inequality is:
\[ x \in (2, \infty) \]