If `log_(3) ( 3 + x) + log_(3) (8 - x) - log_(3) ( 9x - 8) = 2 - log_(3) 9, ` then x =

Options

A. 4
B. `-4`
C. `-8`
D. none of these

Answer

To solve the equation \( \log_3(3 + x) + \log_3(8 - x) - \log_3(9x - 8) = 2 - \log_3(9) \), we can follow these steps:

### Step 1: Combine the logarithms
Using the properties of logarithms, we can combine the left-hand side:
\[ \log_3(3 + x) + \log_3(8 - x) = \log_3((3 + x)(8 - x)) \]
Thus, the equation becomes:
\[ \log_3((3 + x)(8 - x)) - \log_3(9x - 8) = 2 - \log_3(9) \]

### Step 2: Use the quotient property of logarithms
Now, we can apply the quotient property of logarithms:
\[ \log_3\left(\frac{(3 + x)(8 - x)}{9x - 8}\right) = 2 - \log_3(9) \]
We know that \( \log_3(9) = \log_3(3^2) = 2 \), so:
\[ 2 - \log_3(9) = 2 - 2 = 0 \]
Thus, we have:
\[ \log_3\left(\frac{(3 + x)(8 - x)}{9x - 8}\right) = 0 \]

### Step 3: Exponentiate both sides
Exponentiating both sides gives:
\[ \frac{(3 + x)(8 - x)}{9x - 8} = 3^0 = 1 \]

### Step 4: Set up the equation
This leads to:
\[ (3 + x)(8 - x) = 9x - 8 \]

### Step 5: Expand and simplify
Expanding the left side:
\[ 24 - 3x + 8x - x^2 = 9x - 8 \]
This simplifies to:
\[ 24 + 5x - x^2 = 9x - 8 \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ -x^2 + 5x + 24 + 8 - 9x = 0 \]
\[ -x^2 - 4x + 32 = 0 \]
Multiplying through by -1:
\[ x^2 + 4x - 32 = 0 \]

### Step 7: Factor the quadratic
Factoring the quadratic:
\[ (x - 4)(x + 8) = 0 \]
Thus, the solutions are:
\[ x = 4 \quad \text{or} \quad x = -8 \]

### Step 8: Determine valid solutions
Since we are dealing with logarithms, we need to ensure that the arguments of the logarithms are positive:
1. For \( x = 4 \):
- \( 3 + 4 = 7 > 0 \)
- \( 8 - 4 = 4 > 0 \)
- \( 9(4) - 8 = 36 - 8 = 28 > 0 \) (valid)

2. For \( x = -8 \):
- \( 3 - 8 = -5 < 0 \) (not valid)

Thus, the only valid solution is:
\[ \boxed{4} \]