Question: express this as a single logarithm log square root c base c + log square root 12 base 3
1. **State the problem:**
Express $$\log_c \sqrt{c} + \log_3 \sqrt{12}$$ as a single logarithm.
2. **Recall logarithm properties:**
- $$\log_a b^k = k \log_a b$$
- $$\log_a x + \log_a y = \log_a (xy)$$ only if the bases are the same.
3. **Rewrite each term:**
- $$\log_c \sqrt{c} = \log_c c^{1/2} = \frac{1}{2} \log_c c = \frac{1}{2} \times 1 = \frac{1}{2}$$ because $$\log_c c = 1$$.
- $$\log_3 \sqrt{12} = \log_3 12^{1/2} = \frac{1}{2} \log_3 12$$.
4. **Since the bases differ, convert the first term to base 3:**
Use change of base formula:
$$\log_c \sqrt{c} = \frac{\log_3 \sqrt{c}}{\log_3 c} = \frac{\frac{1}{2} \log_3 c}{\log_3 c} = \frac{1}{2}$$
This confirms the first term is $$\frac{1}{2}$$.
5. **Sum the terms:**
$$\frac{1}{2} + \frac{1}{2} \log_3 12 = \frac{1}{2} (1 + \log_3 12)$$.
6. **Rewrite 1 as $$\log_3 3$$:**
$$\frac{1}{2} (\log_3 3 + \log_3 12) = \frac{1}{2} \log_3 (3 \times 12) = \frac{1}{2} \log_3 36$$.
7. **Express as a single logarithm:**
$$\frac{1}{2} \log_3 36 = \log_3 36^{1/2} = \log_3 6$$.
**Final answer:** $$\log_3 6$$.