If ` log_(2) x xx log_(3) x = log_(2) x + log_(3) x`, then find x .

Explanation

To solve the equation \( \log_2 x \cdot \log_3 x = \log_2 x + \log_3 x \), we will follow these steps:

### Step 1: Rewrite the logarithms using the change of base formula
Using the change of base formula, we can express the logarithms as:
\[ \log_2 x = \frac{\log x}{\log 2} \quad \text{and} \quad \log_3 x = \frac{\log x}{\log 3} \]
Substituting these into the equation gives:
\[ \frac{\log x}{\log 2} \cdot \frac{\log x}{\log 3} = \frac{\log x}{\log 2} + \frac{\log x}{\log 3} \]

### Step 2: Simplify the equation
This simplifies to:
\[ \frac{(\log x)^2}{\log 2 \cdot \log 3} = \frac{\log x}{\log 2} + \frac{\log x}{\log 3} \]
We can multiply both sides by \( \log 2 \cdot \log 3 \) to eliminate the denominators:
\[ (\log x)^2 = \log x \cdot \log 3 + \log x \cdot \log 2 \]

### Step 3: Factor out \( \log x \)
Rearranging gives:
\[ (\log x)^2 - \log x \cdot (\log 2 + \log 3) = 0 \]
Factoring out \( \log x \):
\[ \log x \left( \log x - (\log 2 + \log 3) \right) = 0 \]

### Step 4: Set each factor to zero
This gives us two cases to consider:
1. \( \log x = 0 \)
2. \( \log x - (\log 2 + \log 3) = 0 \)

### Step 5: Solve the first case
For the first case:
\[ \log x = 0 \implies x = 10^0 = 1 \]

### Step 6: Solve the second case
For the second case:
\[ \log x = \log 2 + \log 3 \implies \log x = \log(2 \cdot 3) = \log 6 \implies x = 6 \]

### Conclusion
Thus, the solutions are:
\[ x = 1 \quad \text{and} \quad x = 6 \]