Logarithm Simplify 9Be087

1. **State the problem:** Simplify the expression $$\log_c \sqrt{c} + \log_3 \sqrt{12}$$. 2. **Recall the logarithm and root rules:** - $$\log_a \sqrt{x} = \log_a x^{\frac{1}{2}} = \frac{1}{2} \log_a x$$. - The change of base formula: $$\log_a b = \frac{\log_c b}{\log_c a}$$ if needed. 3. **Simplify each term:** - For the first term: $$\log_c \sqrt{c} = \log_c c^{\frac{1}{2}} = \frac{1}{2} \log_c c$$ Since $$\log_c c = 1$$, this becomes $$\frac{1}{2} \times 1 = \frac{1}{2}$$. - For the second term: $$\log_3 \sqrt{12} = \log_3 12^{\frac{1}{2}} = \frac{1}{2} \log_3 12$$. 4. **Rewrite the expression:** $$\frac{1}{2} + \frac{1}{2} \log_3 12 = \frac{1}{2} (1 + \log_3 12)$$. 5. **Simplify $$\log_3 12$$:** - Factor 12 as $$12 = 3 \times 4$$. - Use the product rule: $$\log_3 12 = \log_3 (3 \times 4) = \log_3 3 + \log_3 4 = 1 + \log_3 4$$. 6. **Substitute back:** $$\frac{1}{2} (1 + 1 + \log_3 4) = \frac{1}{2} (2 + \log_3 4) = 1 + \frac{1}{2} \log_3 4$$. 7. **Final answer:** $$1 + \frac{1}{2} \log_3 4$$. This is the simplified form of the original expression.