Evaluate :
(i) `log_(125) 625 - log_(16) 64`
(ii) `log_(16) 32 - log_(25) 125 + log_(9) 27`.

Answer

Let's evaluate the given logarithmic expressions step by step.

### Part (i): Evaluate \( \log_{125} 625 - \log_{16} 64 \)

1. **Rewrite the logarithms using the change of base formula:**
\[ \log_{125} 625 = \frac{\log 625}{\log 125}, \quad \log_{16} 64 = \frac{\log 64}{\log 16} \]
Thus, we can rewrite the expression:
\[ \log_{125} 625 - \log_{16} 64 = \frac{\log 625}{\log 125} - \frac{\log 64}{\log 16} \]

2. **Express the numbers as powers of their prime factors:**
- \( 125 = 5^3 \) and \( 625 = 5^4 \)
- \( 16 = 2^4 \) and \( 64 = 2^6 \)

3. **Substituting these values into the logarithms:**
\[ \log_{125} 625 = \log_{5^3} 5^4 = \frac{4}{3} \quad \text{and} \quad \log_{16} 64 = \log_{2^4} 2^6 = \frac{6}{4} = \frac{3}{2} \]

4. **Now substitute back into the expression:**
\[ \frac{4}{3} - \frac{3}{2} \]

5. **Find a common denominator (LCM of 3 and 2 is 6):**
\[ \frac{4}{3} = \frac{8}{6}, \quad \frac{3}{2} = \frac{9}{6} \]

6. **Subtract the fractions:**
\[ \frac{8}{6} - \frac{9}{6} = \frac{8 - 9}{6} = \frac{-1}{6} \]

**Final Answer for Part (i):**
\[ \log_{125} 625 - \log_{16} 64 = -\frac{1}{6} \]

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### Part (ii): Evaluate \( \log_{16} 32 - \log_{25} 125 + \log_{9} 27 \)

1. **Rewrite the logarithms using the change of base formula:**
\[ \log_{16} 32 = \frac{\log 32}{\log 16}, \quad \log_{25} 125 = \frac{\log 125}{\log 25}, \quad \log_{9} 27 = \frac{\log 27}{\log 9} \]

2. **Express the numbers as powers of their prime factors:**
- \( 16 = 2^4 \) and \( 32 = 2^5 \)
- \( 25 = 5^2 \) and \( 125 = 5^3 \)
- \( 9 = 3^2 \) and \( 27 = 3^3 \)

3. **Substituting these values into the logarithms:**
\[ \log_{16} 32 = \log_{2^4} 2^5 = \frac{5}{4}, \quad \log_{25} 125 = \log_{5^2} 5^3 = \frac{3}{2}, \quad \log_{9} 27 = \log_{3^2} 3^3 = \frac{3}{2} \]

4. **Now substitute back into the expression:**
\[ \frac{5}{4} - \frac{3}{2} + \frac{3}{2} \]

5. **Notice that \( -\frac{3}{2} + \frac{3}{2} = 0 \):**
\[ \frac{5}{4} + 0 = \frac{5}{4} \]

**Final Answer for Part (ii):**
\[ \log_{16} 32 - \log_{25} 125 + \log_{9} 27 = \frac{5}{4} \]

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