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

Options

A. `(2,oo)`
B. `(1,2)`
C. `(-2,-1)`
D. `(1, (3)/(2))`

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 \( 0.09 = 0.3^2 \). Therefore, we can rewrite the second logarithm:
\[ \log_{0.09}(x - 1) = \log_{0.3^2}(x - 1) = \frac{1}{2} \log_{0.3}(x - 1) \]
Now, we can substitute this back into the original inequality:
\[ \log_{0.3}(x - 1) < \frac{1}{2} \log_{0.3}(x - 1) \]

### Step 2: Simplify the inequality
Next, we can manipulate the inequality:
\[ \log_{0.3}(x - 1) < \frac{1}{2} \log_{0.3}(x - 1) \]
Subtract \( \log_{0.3}(x - 1) \) from both sides:
\[ 0 < \frac{1}{2} \log_{0.3}(x - 1) - \log_{0.3}(x - 1) \]
This simplifies to:
\[ 0 < -\frac{1}{2} \log_{0.3}(x - 1) \]
Multiplying through by -2 (and reversing the inequality):
\[ 0 > \log_{0.3}(x - 1) \]

### Step 3: Solve the logarithmic inequality
The inequality \( \log_{0.3}(x - 1) < 0 \) implies:
\[ x - 1 < 1 \quad \text{(since the base 0.3 is less than 1)} \]
Thus,
\[ x < 2 \]

### Step 4: Determine the domain of the logarithm
For the logarithm \( \log_{0.3}(x - 1) \) to be defined, we need:
\[ x - 1 > 0 \implies x > 1 \]

### Step 5: Combine the results
Now we combine the results from steps 3 and 4:
\[ 1 < x < 2 \]

### Conclusion
Thus, the solution for \( x \) lies in the interval:
\[ \boxed{(1, 2)} \]