If `x^(log_(7)x)gt7` where `x gt 0`. Then what is the domain of x ?

Options

A. `x in (0,oo)`
B. `x in (1//7//7)`,
C. `x in (0.1//7) cup(7,oo)`
D. `x in (1//7,oo)`

Answer

To solve the inequality \( x^{\log_{7}x} > 7 \) where \( x > 0 \), we can follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ x^{\log_{7}x} > 7 \]
We can rewrite \( 7 \) as \( 7^1 \) to facilitate comparison:
\[ x^{\log_{7}x} > 7^1 \]

### Step 2: Take Logarithm on Both Sides
Taking logarithm base 7 on both sides, we have:
\[ \log_{7}(x^{\log_{7}x}) > \log_{7}(7) \]
Using the property of logarithms, this simplifies to:
\[ \log_{7}x \cdot \log_{7}x > 1 \]
or
\[ (\log_{7}x)^2 > 1 \]

### Step 3: Solve the Quadratic Inequality
The inequality \( (\log_{7}x)^2 > 1 \) implies two cases:
1. \( \log_{7}x > 1 \)
2. \( \log_{7}x < -1 \)

#### Case 1: \( \log_{7}x > 1 \)
This means:
\[ x > 7^1 \implies x > 7 \]

#### Case 2: \( \log_{7}x < -1 \)
This means:
\[ x < 7^{-1} \implies x < \frac{1}{7} \]

### Step 4: Combine the Results
From the two cases, we have:
1. \( x > 7 \)
2. \( x < \frac{1}{7} \)

Since \( x > 0 \), we can combine these results to state the domain of \( x \):
\[ x \in (0, \frac{1}{7}) \cup (7, \infty) \]

### Final Answer
Thus, the domain of \( x \) is:
\[ (0, \frac{1}{7}) \cup (7, \infty) \]

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