If `x^(log_(3)x^(2)+(log_(3)x)^(2)-10)=1//x^(2)`, then x=

Options

A. `3`
B. `9`
C. `27`
D. `81`

Answer

To solve the equation \( x^{\log_3 x^2 + (\log_3 x)^2 - 10} = \frac{1}{x^2} \), we can follow these steps:

### Step 1: Rewrite the equation
Start by rewriting the right side of the equation:
\[ x^{\log_3 x^2 + (\log_3 x)^2 - 10} = x^{-2} \]

### Step 2: Set the exponents equal
Since the bases are the same, we can set the exponents equal to each other:
\[ \log_3 x^2 + (\log_3 x)^2 - 10 = -2 \]

### Step 3: Simplify the equation
Rearranging gives us:
\[ \log_3 x^2 + (\log_3 x)^2 = 8 \]
Using the property of logarithms, \( \log_3 x^2 = 2 \log_3 x \), we can substitute:
\[ 2 \log_3 x + (\log_3 x)^2 = 8 \]

### Step 4: Let \( y = \log_3 x \)
Let \( y = \log_3 x \). The equation now becomes:
\[ 2y + y^2 = 8 \]
Rearranging gives:
\[ y^2 + 2y - 8 = 0 \]

### Step 5: Factor the quadratic equation
Now we can factor the quadratic:
\[ (y + 4)(y - 2) = 0 \]
Setting each factor to zero gives:
\[ y + 4 = 0 \quad \text{or} \quad y - 2 = 0 \]
Thus:
\[ y = -4 \quad \text{or} \quad y = 2 \]

### Step 6: Solve for \( x \)
Recall that \( y = \log_3 x \). Therefore:
1. For \( y = -4 \):
\[ \log_3 x = -4 \implies x = 3^{-4} = \frac{1}{81} \]
2. For \( y = 2 \):
\[ \log_3 x = 2 \implies x = 3^2 = 9 \]

### Final Answers
The solutions for \( x \) are:
\[ x = \frac{1}{81} \quad \text{and} \quad x = 9 \]