If `"log"_(3) x xx "log"_(x) 2x xx "log"_(2x)y ="log"_(x) x^(2)`, then y equals

Options

A. 9
B. 18
C. 27
D. 81

Answer

To solve the equation \( \log_{3} x \cdot \log_{x} (2x) \cdot \log_{(2x)} y = \log_{x} (x^{2}) \), we can follow these steps:

### Step 1: Rewrite the logarithms using the change of base formula
Using the change of base formula, we can express each logarithm in terms of natural logarithms (or any other base). The change of base formula states that:

\[ \log_{a} b = \frac{\log b}{\log a} \]

Applying this to our equation:

\[ \log_{3} x = \frac{\log x}{\log 3} \]
\[ \log_{x} (2x) = \frac{\log (2x)}{\log x} = \frac{\log 2 + \log x}{\log x} \]
\[ \log_{(2x)} y = \frac{\log y}{\log (2x)} = \frac{\log y}{\log 2 + \log x} \]
\[ \log_{x} (x^{2}) = \frac{\log (x^{2})}{\log x} = \frac{2 \log x}{\log x} = 2 \]

### Step 2: Substitute these into the original equation
Substituting these into the original equation gives us:

\[ \frac{\log x}{\log 3} \cdot \frac{\log 2 + \log x}{\log x} \cdot \frac{\log y}{\log 2 + \log x} = 2 \]

### Step 3: Simplify the equation
Notice that \( \log x \) in the numerator and denominator cancels out:

\[ \frac{\log 2 + \log x}{\log 3} \cdot \frac{\log y}{\log 2 + \log x} = 2 \]

Now, \( \log 2 + \log x \) in the numerator and denominator also cancels out:

\[ \frac{\log y}{\log 3} = 2 \]

### Step 4: Solve for \( \log y \)
Multiplying both sides by \( \log 3 \):

\[ \log y = 2 \log 3 \]

### Step 5: Exponentiate to find \( y \)
Using the property of logarithms that states \( \log a = b \) implies \( a = 10^{b} \) (or the base you are using):

\[ y = 3^{2} \]

### Step 6: Calculate \( y \)
Thus,

\[ y = 9 \]

### Final Answer
Therefore, the value of \( y \) is \( 9 \).

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