Question: now do log square root 3 base 6 + log square root 24 base 6 - log square root 2 base 6
1. **State the problem:** Simplify the expression $$\log_6 \sqrt{3} + \log_6 \sqrt{24} - \log_6 \sqrt{2}$$.
2. **Recall logarithm properties:**
- $$\log_b (xy) = \log_b x + \log_b y$$
- $$\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y$$
- $$\log_b (x^r) = r \log_b x$$
3. **Rewrite square roots as exponents:**
$$\sqrt{3} = 3^{\frac{1}{2}}, \quad \sqrt{24} = 24^{\frac{1}{2}}, \quad \sqrt{2} = 2^{\frac{1}{2}}$$
4. **Apply the power rule:**
$$\log_6 \sqrt{3} = \frac{1}{2} \log_6 3$$
$$\log_6 \sqrt{24} = \frac{1}{2} \log_6 24$$
$$\log_6 \sqrt{2} = \frac{1}{2} \log_6 2$$
5. **Substitute back into the expression:**
$$\frac{1}{2} \log_6 3 + \frac{1}{2} \log_6 24 - \frac{1}{2} \log_6 2$$
6. **Factor out $$\frac{1}{2}$$:**
$$\frac{1}{2} \left( \log_6 3 + \log_6 24 - \log_6 2 \right)$$
7. **Use log addition and subtraction rules:**
$$\log_6 3 + \log_6 24 = \log_6 (3 \times 24) = \log_6 72$$
So the expression inside parentheses becomes:
$$\log_6 72 - \log_6 2 = \log_6 \left( \frac{72}{2} \right) = \log_6 36$$
8. **Simplify the expression:**
$$\frac{1}{2} \log_6 36$$
9. **Express 36 as a power of 6:**
$$36 = 6^2$$
10. **Apply the power rule again:**
$$\frac{1}{2} \log_6 6^2 = \frac{1}{2} \times 2 = 1$$
**Final answer:** $$1$$