The number of solutions of the equation
`3"log"_(3)|-x| = "log"_(3) x^(2),` is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the equation \( 3 \log_{3} | -x | = \log_{3} x^{2} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 3 \log_{3} | -x | = \log_{3} x^{2} \]
Using the property of logarithms that states \( a \log_b c = \log_b(c^a) \), we can rewrite the left side:
\[ \log_{3} (|-x|^3) = \log_{3} x^{2} \]

### Step 2: Set the arguments equal
Since the logarithms have the same base, we can equate the arguments:
\[ |-x|^3 = x^{2} \]

### Step 3: Simplify the absolute value
The absolute value \( |-x| \) is equal to \( |x| \). Therefore, we can rewrite the equation as:
\[ |x|^3 = x^{2} \]

### Step 4: Consider cases for \( |x| \)
We can analyze this equation by considering two cases: \( x \geq 0 \) and \( x < 0 \).

#### Case 1: \( x \geq 0 \)
In this case, \( |x| = x \), and the equation becomes:
\[ x^3 = x^2 \]
Factoring out \( x^2 \):
\[ x^2 (x - 1) = 0 \]
This gives us the solutions:
\[ x^2 = 0 \quad \Rightarrow \quad x = 0 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

#### Case 2: \( x < 0 \)
In this case, \( |x| = -x \), and the equation becomes:
\[ (-x)^3 = x^2 \]
This simplifies to:
\[ -x^3 = x^2 \]
Factoring out \( x^2 \):
\[ x^2 (-x - 1) = 0 \]
This gives us the solutions:
\[ x^2 = 0 \quad \Rightarrow \quad x = 0 \]
\[ -x - 1 = 0 \quad \Rightarrow \quad x = -1 \]

### Step 5: Collect all solutions
From both cases, we have the potential solutions:
- From Case 1: \( x = 0 \) and \( x = 1 \)
- From Case 2: \( x = -1 \)

### Step 6: Eliminate invalid solutions
The logarithm \( \log_{3} x^{2} \) is defined only for \( x \neq 0 \). Thus, we discard \( x = 0 \).

### Final Solutions
The valid solutions are:
- \( x = 1 \)
- \( x = -1 \)

Thus, the total number of solutions to the equation is **2**.