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 \( \log_e(x^2 + 15a^2) - \log_e(a - 2) = \log_e\left(\frac{8ax}{a - 2}\right) \) given that \( x = 9 \) is one of the solutions, we can follow these steps:

### Step 1: Apply the Logarithmic Property
Using the property of logarithms that states \( \log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right) \), we can rewrite the left-hand side of the equation:

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

### Step 2: Eliminate the Logarithm
Since the logarithms are equal, we can set the arguments equal to each other (assuming \( a - 2 > 0 \)):

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

### Step 3: Cross-Multiply
We can eliminate the denominators by cross-multiplying:

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

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

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

### Step 5: Factor the Quadratic Equation
Next, we factor the quadratic equation:

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

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

1. \( x - 3a = 0 \) → \( x = 3a \)
2. \( x - 5a = 0 \) → \( x = 5a \)

### Step 7: Substitute \( x = 9 \)
Since we know \( x = 9 \), we can substitute this value into both equations:

1. From \( x = 3a \):
\[ 9 = 3a \implies a = 3 \]

2. From \( x = 5a \):
\[ 9 = 5a \implies a = \frac{9}{5} \]

### Step 8: Check Validity of Solutions
We need to check the validity of both values of \( a \) with respect to the condition \( a - 2 > 0 \):

1. For \( a = 3 \):
\[ a - 2 = 3 - 2 = 1 > 0 \quad \text{(valid)} \]

2. For \( a = \frac{9}{5} \):
\[ a - 2 = \frac{9}{5} - 2 = \frac{9}{5} - \frac{10}{5} = -\frac{1}{5} < 0 \quad \text{(invalid)} \]

Thus, the only valid solution is \( a = 3 \).

### Conclusion
The solution to the problem is \( a = 3 \).

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