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.
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Break down each term:
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The first term: \[ \frac{1}{2} \log \frac{25}{4} = \log \left( \frac{25}{4} \right)^{1/2} = \log \frac{5}{2} \] (since \( \left( \frac{25}{4} \right)^{1/2} = \frac{5}{2} \))
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The second term: \[ -2 \log \frac{4}{5} = \log \left( \frac{4}{5} \right)^{-2} = \log \frac{25}{16} \] (since \( \left( \frac{4}{5} \right)^{-2} = \frac{5^2}{4^2} = \frac{25}{16} \))
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The third term is already in the logarithmic form: \[ \log \frac{320}{125} \]
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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: \[ \log \left( \frac{5}{2} \cdot \frac{25}{16} \cdot \frac{320}{125} \right) \]
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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: \[ \frac{40000}{4000} = 10 \]
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Final expression: Combining gives us: \[ \log 10 \] Since \( \log 10 = 1 \) (assuming base 10).
Thus, the final answer is: \[ \boxed{1} \]