How many values of `(xgt1)` satisfy the following equation:
`log_(2)xxlog_(4)xxlog_(6)x=log_(2)x.log_(4)x+log_(2)x.log_(6)x+log_(4)x.log_(6)x` ?

Options

A. 0
B. 1
C. 2
D. More than 2

Answer

To solve the equation \( \log_2 x \cdot \log_4 x \cdot \log_6 x = \log_2 x \cdot \log_4 x + \log_2 x \cdot \log_6 x + \log_4 x \cdot \log_6 x \), we will follow these steps:

### Step 1: Rewrite the logarithms
We can express \( \log_4 x \) and \( \log_6 x \) in terms of \( \log_2 x \):
\[ \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2} \]
\[ \log_6 x = \frac{\log_2 x}{\log_2 6} \]

### Step 2: Substitute into the equation
Substituting these into the original equation gives:
\[ \log_2 x \cdot \left(\frac{\log_2 x}{2}\right) \cdot \left(\frac{\log_2 x}{\log_2 6}\right) = \log_2 x \cdot \left(\frac{\log_2 x}{2}\right) + \log_2 x \cdot \left(\frac{\log_2 x}{\log_2 6}\right) + \left(\frac{\log_2 x}{2}\right) \cdot \left(\frac{\log_2 x}{\log_2 6}\right) \]

### Step 3: Simplify both sides
The left-hand side simplifies to:
\[ \frac{(\log_2 x)^3}{2 \log_2 6} \]
The right-hand side simplifies to:
\[ \frac{(\log_2 x)^2}{2} + \frac{(\log_2 x)^2}{\log_2 6} + \frac{(\log_2 x)^2}{2 \log_2 6} \]
Factoring out \( \log_2 x \) gives:
\[ \log_2 x \left(\frac{\log_2 x}{2} + \frac{\log_2 x}{\log_2 6} + \frac{\log_2 x}{2 \log_2 6}\right) \]

### Step 4: Set the equation
Now we can set the two sides equal:
\[ \frac{(\log_2 x)^3}{2 \log_2 6} = \log_2 x \left(\frac{\log_2 x}{2} + \frac{\log_2 x}{\log_2 6} + \frac{\log_2 x}{2 \log_2 6}\right) \]

### Step 5: Cancel \( \log_2 x \) (assuming \( \log_2 x \neq 0 \))
Dividing both sides by \( \log_2 x \) gives:
\[ \frac{(\log_2 x)^2}{2 \log_2 6} = \frac{\log_2 x}{2} + \frac{\log_2 x}{\log_2 6} + \frac{\log_2 x}{2 \log_2 6} \]

### Step 6: Rearranging the equation
Multiplying through by \( 2 \log_2 6 \) to eliminate the fraction:
\[ (\log_2 x)^2 = \log_2 6 \cdot \log_2 x + 2 \log_2 x + \log_2 x \]
This simplifies to:
\[ (\log_2 x)^2 - (3 + \log_2 6) \log_2 x = 0 \]

### Step 7: Factor the quadratic
Factoring gives:
\[ \log_2 x \left(\log_2 x - (3 + \log_2 6)\right) = 0 \]
This gives two solutions:
1. \( \log_2 x = 0 \) (not valid since \( x > 1 \))
2. \( \log_2 x = 3 + \log_2 6 \)

### Step 8: Solve for \( x \)
From the second equation:
\[ \log_2 x = \log_2 (6 \cdot 8) = \log_2 48 \]
Thus, \( x = 48 \).

### Conclusion
Since \( x = 48 \) is the only solution greater than 1, the answer is:
**1 value of \( x \) satisfies the equation.**