`x^(log_(9)x)gt9` implies

Options

A. `xepsilon(0,oo)`
B. `xepsilon(0,(1)/(9))uu(9,oo)`
C. `xgt1`
D. `xlt2`

Answer

To solve the inequality \( x^{\log_{9} x} > 9 \), we can follow these steps:

### Step 1: Rewrite the inequality
We start with the inequality:
\[ x^{\log_{9} x} > 9 \]

### Step 2: Express 9 in terms of base 9
We know that \( 9 = 9^1 \). Thus, we can rewrite the inequality as:
\[ x^{\log_{9} x} > 9^1 \]

### Step 3: Take logarithm base 9 on both sides
Taking logarithm base 9 on both sides gives us:
\[ \log_{9}(x^{\log_{9} x}) > \log_{9}(9) \]

### Step 4: Simplify using logarithmic properties
Using the property of logarithms, \( \log_{a}(b^c) = c \cdot \log_{a}(b) \), we can simplify the left side:
\[ \log_{9} x \cdot \log_{9} x > 1 \]
This simplifies to:
\[ (\log_{9} x)^2 > 1 \]

### Step 5: Solve the quadratic inequality
Now, we can solve the inequality:
\[ (\log_{9} x)^2 - 1 > 0 \]
This can be factored as:
\[ (\log_{9} x - 1)(\log_{9} x + 1) > 0 \]

### Step 6: Find the critical points
The critical points are found by setting each factor to zero:
1. \( \log_{9} x - 1 = 0 \) implies \( \log_{9} x = 1 \) or \( x = 9 \)
2. \( \log_{9} x + 1 = 0 \) implies \( \log_{9} x = -1 \) or \( x = \frac{1}{9} \)

### Step 7: Test intervals
We test the intervals determined by the critical points \( x = \frac{1}{9} \) and \( x = 9 \):
- For \( x < \frac{1}{9} \): Choose \( x = \frac{1}{10} \)
- \( \log_{9} \frac{1}{10} < -1 \) (negative)
- For \( \frac{1}{9} < x < 9 \): Choose \( x = 1 \)
- \( \log_{9} 1 = 0 \) (zero)
- For \( x > 9 \): Choose \( x = 10 \)
- \( \log_{9} 10 > 1 \) (positive)

### Step 8: Determine the solution set
The inequality \( (\log_{9} x - 1)(\log_{9} x + 1) > 0 \) holds in the intervals:
- \( x < \frac{1}{9} \) and \( x > 9 \)

### Conclusion
Thus, the solution to the inequality \( x^{\log_{9} x} > 9 \) is:
\[ x < \frac{1}{9} \quad \text{or} \quad x > 9 \]