1/2 log 25/4 _ 2 log 4/5 + log 320/125

To solve the expression \( \frac{1}{2} \log \frac{25}{4} - 2 \log \frac{4}{5} + \log \frac{320}{125} \), we will use properties of logarithms.

1. **Break down each term**:
- The first term:
\[
\frac{1}{2} \log \frac{25}{4} = \log \left( \frac{25}{4} \right)^{1/2} = \log \frac{5}{2}
\]
(since \( \left( \frac{25}{4} \right)^{1/2} = \frac{5}{2} \))

- The second term:
\[
-2 \log \frac{4}{5} = \log \left( \frac{4}{5} \right)^{-2} = \log \frac{25}{16}
\]
(since \( \left( \frac{4}{5} \right)^{-2} = \frac{5^2}{4^2} = \frac{25}{16} \))

- The third term is already in the logarithmic form:
\[
\log \frac{320}{125}
\]

2. **Combine the logarithms**:
Now we can combine the three logarithmic terms:
\[
\log \frac{5}{2} + \log \frac{25}{16} + \log \frac{320}{125}
\]
Using the property \( \log a + \log b + \log c = \log (abc) \), we have:
\[
\log \left( \frac{5}{2} \cdot \frac{25}{16} \cdot \frac{320}{125} \right)
\]

3. **Calculate the product**:
Let's calculate \( \frac{5}{2} \cdot \frac{25}{16} \cdot \frac{320}{125} \):

First, simplify each fraction:
- \( \frac{5 \times 25 \times 320}{2 \times 16 \times 125} \)

Now calculate the numerator and denominator:
- Numerator: \( 5 \times 25 = 125 \), then \( 125 \times 320 = 40000 \)
- Denominator: \( 2 \times 16 = 32 \), then \( 32 \times 125 = 4000 \)

Now combine:
\[
\frac{40000}{4000} = 10
\]

4. **Final expression**:
Combining gives us:
\[
\log 10
\]
Since \( \log 10 = 1 \) (assuming base 10).

Thus, the final answer is:
\[
\boxed{1}
\]