Express in terms of log 2 and log 3 :
(i) log 36
(ii) log 144
(iii) log 4.5
(iv) `"log"(26)/(51) - "log" (91)/(119)`
(v) `"log"(75)/(16) - "2 log" (5)/(9) + "log" (32)/(243)`

Answer

To express the given logarithmic expressions in terms of log 2 and log 3, we will use the properties of logarithms. Here is the step-by-step solution for each part of the question:

### (i) log 36

1. **Rewrite 36**:
\[ 36 = 6^2 \]
Therefore,
\[ \log 36 = \log(6^2) \]

2. **Apply the power rule**:
\[ \log(6^2) = 2 \cdot \log 6 \]

3. **Rewrite 6 in terms of 2 and 3**:
\[ 6 = 2 \cdot 3 \implies \log 6 = \log(2 \cdot 3) = \log 2 + \log 3 \]

4. **Substitute back**:
\[ 2 \cdot \log 6 = 2(\log 2 + \log 3) = 2 \log 2 + 2 \log 3 \]

Thus,
\[ \log 36 = 2 \log 2 + 2 \log 3 \]

### (ii) log 144

1. **Rewrite 144**:
\[ 144 = 12^2 \]
Therefore,
\[ \log 144 = \log(12^2) \]

2. **Apply the power rule**:
\[ \log(12^2) = 2 \cdot \log 12 \]

3. **Rewrite 12 in terms of 2 and 3**:
\[ 12 = 4 \cdot 3 = 2^2 \cdot 3 \implies \log 12 = \log(2^2 \cdot 3) = \log(2^2) + \log 3 = 2 \log 2 + \log 3 \]

4. **Substitute back**:
\[ 2 \cdot \log 12 = 2(2 \log 2 + \log 3) = 4 \log 2 + 2 \log 3 \]

Thus,
\[ \log 144 = 4 \log 2 + 2 \log 3 \]

### (iii) log 4.5

1. **Rewrite 4.5**:
\[ 4.5 = \frac{9}{2} \]
Therefore,
\[ \log 4.5 = \log\left(\frac{9}{2}\right) \]

2. **Apply the quotient rule**:
\[ \log\left(\frac{9}{2}\right) = \log 9 - \log 2 \]

3. **Rewrite 9**:
\[ 9 = 3^2 \implies \log 9 = 2 \log 3 \]

4. **Substitute back**:
\[ \log 4.5 = 2 \log 3 - \log 2 \]

Thus,
\[ \log 4.5 = 2 \log 3 - \log 2 \]

### (iv) log(26/51) - log(91/119)

1. **Combine the logarithms**:
\[ \log\left(\frac{26}{51}\right) - \log\left(\frac{91}{119}\right) = \log\left(\frac{26}{51} \cdot \frac{119}{91}\right) \]

2. **Simplify the fraction**:
\[ = \log\left(\frac{26 \cdot 119}{51 \cdot 91}\right) \]

3. **Factor the numbers**:
- \(26 = 2 \cdot 13\)
- \(51 = 3 \cdot 17\)
- \(91 = 7 \cdot 13\)
- \(119 = 7 \cdot 17\)

4. **Substitute and simplify**:
\[ = \log\left(\frac{2 \cdot 13 \cdot 119}{3 \cdot 17 \cdot 91}\right) = \log\left(\frac{2}{3}\right) \]

5. **Rewrite in terms of log 2 and log 3**:
\[ \log\left(\frac{2}{3}\right) = \log 2 - \log 3 \]

Thus,
\[ \log\left(\frac{26}{51}\right) - \log\left(\frac{91}{119}\right) = \log 2 - \log 3 \]

### (v) log(75/16) - 2 log(5/9) + log(32/243)

1. **Rewrite each term**:
\[ \log\left(\frac{75}{16}\right) - 2 \log\left(\frac{5}{9}\right) + \log\left(\frac{32}{243}\right) \]

2. **Apply properties of logarithms**:
\[ = \log\left(\frac{75}{16}\right) - \log\left(\frac{5^2}{9^2}\right) + \log\left(\frac{32}{243}\right) \]

3. **Combine the logarithms**:
\[ = \log\left(\frac{75 \cdot 9^2 \cdot 32}{16 \cdot 25}\right) \]

4. **Factor the numbers**:
- \(75 = 3 \cdot 5^2\)
- \(16 = 2^4\)
- \(32 = 2^5\)
- \(243 = 3^5\)

5. **Substitute and simplify**:
\[ = \log\left(\frac{3 \cdot 5^2 \cdot 9^2 \cdot 2^5}{2^4 \cdot 5^2 \cdot 3^5}\right) = \log\left(\frac{3 \cdot 2^5}{2^4 \cdot 3^5}\right) = \log\left(\frac{2}{3^4}\right) \]

6. **Rewrite in terms of log 2 and log 3**:
\[ = \log 2 - 4 \log 3 \]

Thus,
\[ \log\left(\frac{75}{16}\right) - 2 \log\left(\frac{5}{9}\right) + \log\left(\frac{32}{243}\right) = \log 2 - 4 \log 3 \]

### Summary of Answers:
1. \( \log 36 = 2 \log 2 + 2 \log 3 \)
2. \( \log 144 = 4 \log 2 + 2 \log 3 \)
3. \( \log 4.5 = 2 \log 3 - \log 2 \)
4. \( \log\left(\frac{26}{51}\right) - \log\left(\frac{91}{119}\right) = \log 2 - \log 3 \)
5. \( \log\left(\frac{75}{16}\right) - 2 \log\left(\frac{5}{9}\right) + \log\left(\frac{32}{243}\right) = \log 2 - 4 \log 3 \)