Solve : `log_3 (3^(x)-8)=2-x`

Answer

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

### Step 1: Rewrite the logarithmic equation in exponential form
Using the property of logarithms, we can rewrite the equation:
\[ \log_3(3^x - 8) = 2 - x \implies 3^{2 - x} = 3^x - 8 \]

### Step 2: Simplify the exponential equation
We can express \( 3^{2 - x} \) as:
\[ 3^{2 - x} = \frac{9}{3^x} \]
So, the equation becomes:
\[ \frac{9}{3^x} = 3^x - 8 \]

### Step 3: Multiply both sides by \( 3^x \) to eliminate the fraction
Multiplying both sides by \( 3^x \) gives:
\[ 9 = (3^x)(3^x) - 8(3^x) \]
This simplifies to:
\[ 9 = 3^{2x} - 8 \cdot 3^x \]

### Step 4: Rearrange the equation
Rearranging the equation leads to:
\[ 3^{2x} - 8 \cdot 3^x - 9 = 0 \]

### Step 5: Let \( y = 3^x \)
Substituting \( y = 3^x \) transforms the equation into a quadratic form:
\[ y^2 - 8y - 9 = 0 \]

### Step 6: Factor the quadratic equation
Next, we will factor the quadratic:
\[ (y - 9)(y + 1) = 0 \]

### Step 7: Solve for \( y \)
Setting each factor to zero gives:
\[ y - 9 = 0 \quad \text{or} \quad y + 1 = 0 \]
Thus, we find:
\[ y = 9 \quad \text{or} \quad y = -1 \]

### Step 8: Substitute back to find \( x \)
Since \( y = 3^x \), we have:
1. \( 3^x = 9 \) which implies \( x = 2 \)
2. \( 3^x = -1 \) is not possible since \( 3^x \) cannot be negative.

### Conclusion
The only solution is:
\[ x = 2 \]