If ` log_(5) x = a and log_(2) y = a ," find "100^(2a-1)` in terms of x and y .

Explanation

To solve the problem, we need to find \( 100^{(2a-1)} \) in terms of \( x \) and \( y \), given that \( \log_5 x = a \) and \( \log_2 y = a \).

### Step-by-Step Solution:

1. **Convert logarithmic equations to exponential form:**
- From \( \log_5 x = a \), we can express \( x \) as:
\[ x = 5^a \]
- From \( \log_2 y = a \), we can express \( y \) as:
\[ y = 2^a \]

2. **Express \( 100 \) in terms of its prime factors:**
- We know that:
\[ 100 = 10^2 = (2 \cdot 5)^2 = 2^2 \cdot 5^2 \]

3. **Substitute \( 100 \) into the expression \( 100^{(2a-1)} \):**
- We can rewrite \( 100^{(2a-1)} \) as:
\[ 100^{(2a-1)} = (2^2 \cdot 5^2)^{(2a-1)} \]

4. **Apply the power of a product property:**
- Using the property \( (ab)^m = a^m \cdot b^m \), we have:
\[ (2^2 \cdot 5^2)^{(2a-1)} = 2^{2(2a-1)} \cdot 5^{2(2a-1)} \]

5. **Simplify the exponents:**
- This simplifies to:
\[ 2^{4a-2} \cdot 5^{4a-2} \]

6. **Factor out the common exponent:**
- We can factor this as:
\[ 2^{4a} \cdot 5^{4a} \cdot \frac{1}{2^2 \cdot 5^2} = (2^a \cdot 5^a)^4 \cdot \frac{1}{100} \]

7. **Substitute back the expressions for \( x \) and \( y \):**
- Since \( 2^a = y \) and \( 5^a = x \), we have:
\[ (xy)^4 \cdot \frac{1}{100} \]

8. **Final expression:**
- Thus, we can conclude:
\[ 100^{(2a-1)} = \frac{x^4 y^4}{100} \]

### Final Answer:
\[ 100^{(2a-1)} = \frac{x^4 y^4}{100} \]