If `log_(1//sqrt(2))(x-1)gt2` then `x` lies in the interval

Options

A. `((3)/(2),oo)`
B. `(-oo,(3)/(2))`
C. `((3)/(2),1)`
D. none of these

Answer

To solve the inequality \( \log_{\frac{1}{\sqrt{2}}}(x-1) > 2 \), we can follow these steps:

### Step 1: Understand the logarithmic inequality
The logarithmic inequality \( \log_a(b) > c \) can be rewritten based on the base \( a \):
- If \( 0 < a < 1 \), then \( \log_a(b) > c \) implies \( b < a^c \).

In our case, the base \( \frac{1}{\sqrt{2}} \) is less than 1. Therefore, we can rewrite the inequality as:
\[ x - 1 < \left(\frac{1}{\sqrt{2}}\right)^2 \]

### Step 2: Calculate \( \left(\frac{1}{\sqrt{2}}\right)^2 \)
Calculating the right side:
\[ \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
Thus, the inequality becomes:
\[ x - 1 < \frac{1}{2} \]

### Step 3: Solve for \( x \)
Now, we solve for \( x \):
\[ x < 1 + \frac{1}{2} \]
\[ x < \frac{3}{2} \]

### Step 4: Consider the domain of the logarithm
Since we have \( \log_{\frac{1}{\sqrt{2}}}(x-1) \), the argument \( x - 1 \) must be greater than 0:
\[ x - 1 > 0 \implies x > 1 \]

### Step 5: Combine the inequalities
Now we combine the two inequalities:
1. \( x > 1 \)
2. \( x < \frac{3}{2} \)

Thus, we conclude that:
\[ 1 < x < \frac{3}{2} \]

### Final Answer
The solution for \( x \) lies in the interval:
\[ (1, \frac{3}{2}) \]

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