1. **State the problem:** Given the equation $\log y = 3\log 2 + \log 3 - \log 6$, find the value of $y$. 2. **Recall logarithm properties:** - $\log a + \log b = \log(ab)$ - $\log a - \log b = \log\left(\frac{a}{b}\right)$ - $k \log a = \log(a^k)$ 3. **Apply the properties:** Rewrite the right side: $$3\log 2 + \log 3 - \log 6 = \log(2^3) + \log 3 - \log 6 = \log(8) + \log 3 - \log 6$$ 4. **Combine the logarithms:** $$\log(8) + \log 3 = \log(8 \times 3) = \log(24)$$ So, $$\log y = \log(24) - \log 6 = \log\left(\frac{24}{6}\right) = \log 4$$ 5. **Conclude the value of $y$:** Since $\log y = \log 4$, it follows that $$y = 4$$ **Final answer:** $y = 4$