The set of real values of x for which
`2^("log"_(sqrt(2))(x-1)) gt x+ 5,` is

Options

A. `(-oo, -1) cup (4, oo)`
B. `(4, oo)`
C. `(-1, 4)`
D. none of these

Explanation

To solve the inequality \( 2^{\log_{\sqrt{2}}(x-1)} > x + 5 \), we will follow these steps:

### Step 1: Rewrite the logarithm
Using the property of logarithms, we can rewrite \( \log_{\sqrt{2}}(x-1) \):
\[ \log_{\sqrt{2}}(x-1) = \frac{\log_{2}(x-1)}{\log_{2}(\sqrt{2})} = \frac{\log_{2}(x-1)}{1/2} = 2\log_{2}(x-1) \]
Thus, we can rewrite the left side of the inequality:
\[ 2^{\log_{\sqrt{2}}(x-1)} = 2^{2\log_{2}(x-1)} = (2^{\log_{2}(x-1)})^2 = (x-1)^2 \]
So the inequality becomes:
\[ (x-1)^2 > x + 5 \]

### Step 2: Rearrange the inequality
Now we will rearrange the inequality:
\[ (x-1)^2 - (x + 5) > 0 \]
Expanding \( (x-1)^2 \):
\[ x^2 - 2x + 1 - x - 5 > 0 \]
This simplifies to:
\[ x^2 - 3x - 4 > 0 \]

### Step 3: Factor the quadratic
Next, we will factor the quadratic expression:
\[ x^2 - 3x - 4 = (x - 4)(x + 1) \]
Thus, the inequality can be rewritten as:
\[ (x - 4)(x + 1) > 0 \]

### Step 4: Determine the intervals
To find the intervals where this product is positive, we need to find the critical points:
- \( x - 4 = 0 \) gives \( x = 4 \)
- \( x + 1 = 0 \) gives \( x = -1 \)

We will test the intervals defined by these critical points: \( (-\infty, -1) \), \( (-1, 4) \), and \( (4, \infty) \).

1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \):
\[ (-2 - 4)(-2 + 1) = (-6)(-1) = 6 > 0 \quad \text{(True)} \]

2. **Interval \( (-1, 4) \)**: Choose \( x = 0 \):
\[ (0 - 4)(0 + 1) = (-4)(1) = -4 < 0 \quad \text{(False)} \]

3. **Interval \( (4, \infty) \)**: Choose \( x = 5 \):
\[ (5 - 4)(5 + 1) = (1)(6) = 6 > 0 \quad \text{(True)} \]

### Step 5: Combine the intervals
The solution to the inequality \( (x - 4)(x + 1) > 0 \) is:
\[ x \in (-\infty, -1) \cup (4, \infty) \]

### Step 6: Consider the domain of the logarithm
Since \( \log_{\sqrt{2}}(x-1) \) requires \( x-1 > 0 \) or \( x > 1 \), we must consider this restriction. Therefore, we discard the interval \( (-\infty, -1) \).

### Final Solution
The valid solution set is:
\[ x \in (4, \infty) \]