If `(log)_(0. 3)(x-1)<(log)_(0. 09)(x-1),` then `x` lies in the interval `(2,oo)` (b) `(1,2)` `(-2,-1)` (d) None of these

Options

A. `(-infty.1)`
B. (1,2)
C. `(2,infty)`
D. None of the above

Answer

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

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

### Step 2: Use the property of logarithms
Using the property of logarithms that states \( \log_{a^2}(b) = \frac{1}{2} \log_a(b) \), we can rewrite the right-hand side:
\[ \log_{0.3}(x-1) < \frac{1}{2} \log_{0.3}(x-1) \]

### Step 3: Cross-multiply
To eliminate the logarithm, we can cross-multiply:
\[ 2 \log_{0.3}(x-1) < \log_{0.3}(x-1) \]

### Step 4: Rearrange the inequality
Rearranging gives us:
\[ 2 \log_{0.3}(x-1) - \log_{0.3}(x-1) < 0 \]
which simplifies to:
\[ \log_{0.3}(x-1) < 0 \]

### Step 5: Interpret the logarithmic inequality
The inequality \( \log_{0.3}(x-1) < 0 \) implies that \( x-1 < 1 \) because the logarithm of a number less than 1 is negative. Therefore:
\[ x - 1 < 1 \implies x < 2 \]

### Step 6: Determine the valid range for \( x \)
However, we also need \( x-1 > 0 \) for the logarithm to be defined, which gives us:
\[ x - 1 > 0 \implies x > 1 \]

### Step 7: Combine the inequalities
Combining \( x < 2 \) and \( x > 1 \), we find:
\[ 1 < x < 2 \]

### Conclusion
Thus, the solution to the inequality \( \log_{0.3}(x-1) < \log_{0.09}(x-1) \) is that \( x \) lies in the interval \( (1, 2) \).

### Final Answer
The correct option is (b) \( (1, 2) \).

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