If `log_4A = log_6B = log_9(A+B)` then the value of `B/A` is

Options

A. `(sqrt(5)-1)/(4)`
B. `(sqrt(5)+1)/(4)`
C. `(sqrt(5)-1)/(2)`
D. `(sqrt(5)+1)/(2)`

Explanation

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

\[ \log_4 A = \log_6 B = \log_9 (A + B) \]

Let us denote this common value as \( k \). Therefore, we can express \( A \), \( B \), and \( A + B \) in terms of \( k \):

1. **Express \( A \) in terms of \( k \)**:
\[ \log_4 A = k \implies A = 4^k = (2^2)^k = 2^{2k} \]

2. **Express \( B \) in terms of \( k \)**:
\[ \log_6 B = k \implies B = 6^k = (2 \cdot 3)^k = 2^k \cdot 3^k \]

3. **Express \( A + B \) in terms of \( k \)**:
\[ \log_9 (A + B) = k \implies A + B = 9^k = (3^2)^k = 3^{2k} \]

Now we have:
- \( A = 2^{2k} \)
- \( B = 2^k \cdot 3^k \)
- \( A + B = 3^{2k} \)

Next, we can substitute \( A \) and \( B \) into the equation for \( A + B \):

\[ A + B = 2^{2k} + 2^k \cdot 3^k = 3^{2k} \]

4. **Rearranging the equation**:
\[ 2^{2k} + 2^k \cdot 3^k - 3^{2k} = 0 \]

5. **Let \( x = 2^k \)**. Then \( 3^k = \frac{3^{2k}}{x^2} \):
\[ x^2 + x \cdot \left(\frac{3^k}{x}\right) - 3^{2k} = 0 \]

This simplifies to:
\[ x^2 + 3^k x - 3^{2k} = 0 \]

6. **Using the quadratic formula to solve for \( x \)**:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3^k \pm \sqrt{(3^k)^2 + 4 \cdot 3^{2k}}}{2} \]

7. **Finding \( \frac{B}{A} \)**:
\[ \frac{B}{A} = \frac{2^k \cdot 3^k}{2^{2k}} = \frac{3^k}{2^k} = \left(\frac{3}{2}\right)^k \]

Now, we can substitute \( k \) back into the equation to find \( \frac{B}{A} \).

8. **Final Calculation**:
From the quadratic equation, we can find the roots and determine the positive value for \( \frac{B}{A} \).

The final result is:
\[ \frac{B}{A} = \frac{1 + \sqrt{5}}{2} \]