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 property of logarithms
We know that \( \log_x y = \frac{1}{\log_y x} \). Let's denote \( \log_y x \) as \( t \). Therefore, we can rewrite the equation as:
\[ t + \frac{1}{t} = 7 \]

### Step 2: Multiply through by \( t \)
To eliminate the fraction, multiply both sides by \( t \):
\[ t^2 + 1 = 7t \]

### Step 3: Rearrange the equation
Rearranging gives us a standard quadratic equation:
\[ t^2 - 7t + 1 = 0 \]

### Step 4: Apply the quadratic formula
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -7, c = 1 \):
\[ t = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ t = \frac{7 \pm \sqrt{49 - 4}}{2} \]
\[ t = \frac{7 \pm \sqrt{45}}{2} \]
\[ t = \frac{7 \pm 3\sqrt{5}}{2} \]

### Step 5: Find \( t^2 + \frac{1}{t^2} \)
We need to find \( t^2 + \frac{1}{t^2} \). We can use the identity:
\[ t^2 + \frac{1}{t^2} = \left(t + \frac{1}{t}\right)^2 - 2 \]
Substituting \( t + \frac{1}{t} = 7 \):
\[ t^2 + \frac{1}{t^2} = 7^2 - 2 \]
\[ t^2 + \frac{1}{t^2} = 49 - 2 = 47 \]

### Conclusion
Thus, the value of \( (\log_y x)^2 + (\log_x y)^2 \) is:
\[ \boxed{47} \]