If $$ \frac{3^{2 x-8}}{225}=\frac{5^{3}}{5^{x}} $$, then $$ x= $$

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We are given the equation:

$$
\frac{3^{2x - 8}}{225} = \frac{5^3}{5^x}
$$

Let's simplify both sides step by step.

1. **Simplify the Right Side:**
   
   $$
   \frac{5^3}{5^x} = 5^{3 - x}
   $$

2. **Express 225 in Terms of Its Prime Factors:**
   
   $$
   225 = 15^2 = (3 \times 5)^2 = 3^2 \times 5^2
   $$

3. **Rewrite the Left Side Using Prime Factors:**
   
   $$
   \frac{3^{2x - 8}}{225} = \frac{3^{2x - 8}}{3^2 \times 5^2} = 3^{2x - 8 - 2} \times 5^{-2} = 3^{2x - 10} \times 5^{-2}
   $$

4. **Set the Simplified Forms Equal:**
   
   $$
   3^{2x - 10} \times 5^{-2} = 5^{3 - x}
   $$

5. **Equate the Exponents of Like Bases:**
   
   For the bases to be equal, their exponents must satisfy:
   
   - For base 3:
     $$
     2x - 10 = 0 \quad \Rightarrow \quad 2x = 10 \quad \Rightarrow \quad x = 5
     $$
   
   - For base 5:
     $$
     -2 = 3 - x \quad \Rightarrow \quad x = 5
     $$

Both conditions yield $$ x = 5 $$.

**Answer:** 5
$$ x = 5 $$

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