If `log_(y)x +log_(x)y = 7`, then the value of `(log_(y)x)^(2) +(log_(x)y)^(2)`, is

Options

A. 43
B. 45
C. 47
D. 49

Answer

To solve the equation \( \log_{y} x + \log_{x} y = 7 \) and find the value of \( (\log_{y} x)^{2} + (\log_{x} y)^{2} \), we can follow these steps:

### Step 1: Use the Change of Base Formula
We know that:
\[ \log_{x} y = \frac{1}{\log_{y} x} \]
Let \( a = \log_{y} x \). Then, we can rewrite \( \log_{x} y \) as:
\[ \log_{x} y = \frac{1}{a} \]

### Step 2: Substitute into the Original Equation
Substituting these into the original equation gives:
\[ a + \frac{1}{a} = 7 \]

### Step 3: Multiply through by \( a \)
To eliminate the fraction, multiply the entire equation by \( a \):
\[ a^2 + 1 = 7a \]

### Step 4: Rearrange the Equation
Rearranging this gives us a standard quadratic equation:
\[ a^2 - 7a + 1 = 0 \]

### Step 5: Use the Quadratic Formula
We can solve for \( a \) using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -7, c = 1 \):
\[ a = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{7 \pm \sqrt{49 - 4}}{2} = \frac{7 \pm \sqrt{45}}{2} = \frac{7 \pm 3\sqrt{5}}{2} \]

### Step 6: Calculate \( a^2 + \frac{1}{a^2} \)
Now, we need to find \( a^2 + \frac{1}{a^2} \). We can use the identity:
\[ a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2 \]
Substituting \( a + \frac{1}{a} = 7 \):
\[ a^2 + \frac{1}{a^2} = 7^2 - 2 = 49 - 2 = 47 \]

### Final Answer
Thus, the value of \( (\log_{y} x)^{2} + (\log_{x} y)^{2} \) is:
\[ \boxed{47} \]