The total number of integral value of x satisfying the equation ` x ^(log_(3)x ^(2))-10=1 //x ^(2) ` is

Answer

To solve the equation \( x^{(\log_{3}(x^2))} - 10 = \frac{1}{x^2} \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^{(\log_{3}(x^2))} - 10 = \frac{1}{x^2} \]
We can rewrite \( \log_{3}(x^2) \) using the property of logarithms:
\[ \log_{3}(x^2) = 2 \cdot \log_{3}(x) \]
Thus, we can rewrite the left-hand side:
\[ x^{(2 \cdot \log_{3}(x))} - 10 = \frac{1}{x^2} \]

### Step 2: Simplify the left-hand side
Using the property of exponents, we can express \( x^{(2 \cdot \log_{3}(x))} \) as:
\[ x^{(2 \cdot \log_{3}(x))} = (x^{\log_{3}(x)})^2 = 3^{(\log_{3}(x))^2} \]
However, we can also express it directly as:
\[ x^{(2 \cdot \log_{3}(x))} = 3^{\log_{3}(x^2)} = x^2 \]
So the equation becomes:
\[ x^2 - 10 = \frac{1}{x^2} \]

### Step 3: Multiply through by \( x^2 \)
To eliminate the fraction, we multiply both sides by \( x^2 \):
\[ x^4 - 10x^2 = 1 \]

### Step 4: Rearrange the equation
Rearranging gives us a standard polynomial form:
\[ x^4 - 10x^2 - 1 = 0 \]

### Step 5: Let \( y = x^2 \)
Now, we can substitute \( y = x^2 \). This transforms our equation into:
\[ y^2 - 10y - 1 = 0 \]

### Step 6: Solve the quadratic equation
We can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -10, c = -1 \):
\[ y = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ y = \frac{10 \pm \sqrt{100 + 4}}{2} \]
\[ y = \frac{10 \pm \sqrt{104}}{2} \]
\[ y = \frac{10 \pm 2\sqrt{26}}{2} \]
\[ y = 5 \pm \sqrt{26} \]

### Step 7: Find \( x \) values
Since \( y = x^2 \), we have:
\[ x^2 = 5 + \sqrt{26} \quad \text{or} \quad x^2 = 5 - \sqrt{26} \]
Now, we need to check if these values yield integral values for \( x \).

### Step 8: Analyze the roots
1. For \( x^2 = 5 + \sqrt{26} \):
- This is positive and not a perfect square.
2. For \( x^2 = 5 - \sqrt{26} \):
- Calculate \( 5 - \sqrt{26} \):
- Since \( \sqrt{26} \approx 5.1 \), \( 5 - \sqrt{26} < 0 \), which means no real \( x \) exists.

### Conclusion
Since neither \( 5 + \sqrt{26} \) nor \( 5 - \sqrt{26} \) provides integral solutions for \( x \), the total number of integral values of \( x \) satisfying the original equation is:
\[ \text{Total integral values of } x = 0 \]