If x = 9 is one of the solutions of ` log_(e)(x^(2)+15a^(2))-log_(e)(a-2)=log_(e)((8ax)/(a-2))`,then

Options

A. ` a= 3/5`
B. a = 3
C. x = 15
D. x = 2

Explanation

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

\[ \log_e(x^2 + 15a^2) - \log_e(a - 2) = \log_e\left(\frac{8ax}{a - 2}\right) \]

### Step 1: Apply the properties of logarithms
Using the property of logarithms that states \(\log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right)\), we can rewrite the left-hand side:

\[ \log_e\left(\frac{x^2 + 15a^2}{a - 2}\right) = \log_e\left(\frac{8ax}{a - 2}\right) \]

### Step 2: Set the arguments of the logarithms equal
Since the logarithms are equal, we can set their arguments equal to each other:

\[ \frac{x^2 + 15a^2}{a - 2} = \frac{8ax}{a - 2} \]

### Step 3: Multiply both sides by \(a - 2\)
Assuming \(a - 2 \neq 0\), we can multiply both sides by \(a - 2\):

\[ x^2 + 15a^2 = 8ax \]

### Step 4: Rearrange into a standard quadratic form
Rearranging the equation gives us:

\[ x^2 - 8ax + 15a^2 = 0 \]

### Step 5: Use the quadratic formula or factor the quadratic
This is a quadratic equation in \(x\). We can factor it as follows:

\[ (x - 3a)(x - 5a) = 0 \]

### Step 6: Find the solutions for \(x\)
Setting each factor to zero gives us the solutions:

\[ x - 3a = 0 \quad \Rightarrow \quad x = 3a \]
\[ x - 5a = 0 \quad \Rightarrow \quad x = 5a \]

### Step 7: Substitute \(x = 9\) to find \(a\)
Since we know \(x = 9\) is one of the solutions, we can set each equation equal to 9:

1. For \(x = 3a\):
\[ 3a = 9 \quad \Rightarrow \quad a = 3 \]

2. For \(x = 5a\):
\[ 5a = 9 \quad \Rightarrow \quad a = \frac{9}{5} \]

### Final Result
Thus, the values of \(a\) that correspond to the solutions for \(x\) are:

- \(a = 3\)
- \(a = \frac{9}{5}\)