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 problem, we start with the given equation:

\[ \log_y x + \log_x y = 7 \]

### Step 1: Use the property of logarithms

We know that:

\[ \log_x y = \frac{1}{\log_y x} \]

Let \( a = \log_y x \). Then we can rewrite the equation as:

\[ a + \frac{1}{a} = 7 \]

### Step 2: Multiply through by \( a \)

To eliminate the fraction, multiply both sides by \( a \):

\[ a^2 + 1 = 7a \]

### Step 3: Rearrange the equation

Rearranging gives us a quadratic equation:

\[ a^2 - 7a + 1 = 0 \]

### Step 4: Use the quadratic formula

We can solve for \( a \) using the quadratic formula:

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

Here, \( a = 1, b = -7, c = 1 \):

\[ a = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]

Calculating the discriminant:

\[ = \frac{7 \pm \sqrt{49 - 4}}{2} = \frac{7 \pm \sqrt{45}}{2} = \frac{7 \pm 3\sqrt{5}}{2} \]

### Step 5: Find \( a^2 + \frac{1}{a^2} \)

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:

\[ \boxed{47} \]

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