If log 2 = 0.3010 and log 3 = 0.4771, find the value of :
(i) log 6
(ii) log 5
(iii) `log sqrt(24)`

Answer

To solve the given problem, we will find the values of log 6, log 5, and log √24 using the properties of logarithms and the provided values of log 2 and log 3.

Given:
- log 2 = 0.3010
- log 3 = 0.4771

### (i) Finding log 6

1. **Express 6 in terms of 2 and 3:**
\[ 6 = 2 \times 3 \]

2. **Apply the logarithmic property:**
\[ \log 6 = \log(2 \times 3) = \log 2 + \log 3 \]

3. **Substitute the known values:**
\[ \log 6 = 0.3010 + 0.4771 \]

4. **Calculate the sum:**
\[ \log 6 = 0.7781 \]

### (ii) Finding log 5

1. **Express 5 in terms of 10 and 2:**
\[ 5 = \frac{10}{2} \]

2. **Apply the logarithmic property:**
\[ \log 5 = \log(10/2) = \log 10 - \log 2 \]

3. **Substitute the known values:**
\[ \log 10 = 1 \quad \text{(since log base 10 of 10 is 1)} \]
\[ \log 5 = 1 - 0.3010 \]

4. **Calculate the difference:**
\[ \log 5 = 0.6990 \]

### (iii) Finding log √24

1. **Express √24 in terms of its factors:**
\[ \sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2 \times \sqrt{6} \]

2. **Apply the logarithmic property:**
\[ \log \sqrt{24} = \log(2 \times \sqrt{6}) = \log 2 + \log \sqrt{6} \]

3. **Use the property of logarithms for square roots:**
\[ \log \sqrt{6} = \frac{1}{2} \log 6 \]

4. **Substitute the value of log 6:**
\[ \log \sqrt{6} = \frac{1}{2} \times 0.7781 = 0.38905 \]

5. **Combine the results:**
\[ \log \sqrt{24} = \log 2 + \log \sqrt{6} = 0.3010 + 0.38905 \]

6. **Calculate the sum:**
\[ \log \sqrt{24} = 0.69005 \]

### Summary of Results:
- (i) log 6 = 0.7781
- (ii) log 5 = 0.6990
- (iii) log √24 = 0.69005