If `alpha` and `beta` are roots of equation `(log_(3)x)^(2)-4(log_(3)x)+2=0` then
The value of `log_(alpha)beta+log_(beta)alpha` is

Answer

To solve the problem, we start with the given quadratic equation:

\[ (\log_{3} x)^{2} - 4(\log_{3} x) + 2 = 0 \]

Let \( y = \log_{3} x \). Then, we can rewrite the equation as:

\[ y^{2} - 4y + 2 = 0 \]

### Step 1: Solve the quadratic equation
We can use the quadratic formula to find the roots of the equation \( y^{2} - 4y + 2 = 0 \). The quadratic formula is given by:

\[ y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -4 \), and \( c = 2 \). Plugging in these values:

\[ y = \frac{4 \pm \sqrt{(-4)^{2} - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \]

Calculating the discriminant:

\[ b^{2} - 4ac = 16 - 8 = 8 \]

Now substituting back into the formula:

\[ y = \frac{4 \pm \sqrt{8}}{2} \]

Since \( \sqrt{8} = 2\sqrt{2} \), we have:

\[ y = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2} \]

Thus, the roots \( y_1 \) and \( y_2 \) are:

\[ y_1 = 2 + \sqrt{2}, \quad y_2 = 2 - \sqrt{2} \]

### Step 2: Find \( \alpha \) and \( \beta \)
Since \( y = \log_{3} x \), we can express \( x \) in terms of \( \alpha \) and \( \beta \):

\[ \alpha = 3^{y_1} = 3^{2 + \sqrt{2}}, \quad \beta = 3^{y_2} = 3^{2 - \sqrt{2}} \]

### Step 3: Calculate \( \log_{\alpha} \beta + \log_{\beta} \alpha \)
Using the change of base formula, we can express \( \log_{\alpha} \beta \) and \( \log_{\beta} \alpha \):

\[ \log_{\alpha} \beta = \frac{\log_{3} \beta}{\log_{3} \alpha}, \quad \log_{\beta} \alpha = \frac{\log_{3} \alpha}{\log_{3} \beta} \]

Now, we can calculate \( \log_{3} \alpha \) and \( \log_{3} \beta \):

\[ \log_{3} \alpha = y_1 = 2 + \sqrt{2}, \quad \log_{3} \beta = y_2 = 2 - \sqrt{2} \]

Substituting these into the expressions:

\[ \log_{\alpha} \beta = \frac{2 - \sqrt{2}}{2 + \sqrt{2}}, \quad \log_{\beta} \alpha = \frac{2 + \sqrt{2}}{2 - \sqrt{2}} \]

### Step 4: Combine the two logarithmic expressions
Now, we need to find:

\[ \log_{\alpha} \beta + \log_{\beta} \alpha = \frac{2 - \sqrt{2}}{2 + \sqrt{2}} + \frac{2 + \sqrt{2}}{2 - \sqrt{2}} \]

To simplify this, we will find a common denominator:

\[ = \frac{(2 - \sqrt{2})^2 + (2 + \sqrt{2})^2}{(2 + \sqrt{2})(2 - \sqrt{2})} \]

Calculating the numerator:

\[ (2 - \sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2} \]
\[ (2 + \sqrt{2})^2 = 4 + 4\sqrt{2} + 2 = 6 + 4\sqrt{2} \]

Adding these:

\[ (6 - 4\sqrt{2}) + (6 + 4\sqrt{2}) = 12 \]

Calculating the denominator:

\[ (2 + \sqrt{2})(2 - \sqrt{2}) = 4 - 2 = 2 \]

Thus, we have:

\[ \log_{\alpha} \beta + \log_{\beta} \alpha = \frac{12}{2} = 6 \]

### Final Answer
The value of \( \log_{\alpha} \beta + \log_{\beta} \alpha \) is \( 6 \).

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