1. **State the problem:** Simplify the expression $\log 64 - \log 6 - \log 8 + \log \frac{3}{4}$.\n\n2. **Recall logarithm rules:**\n- $\log a - \log b = \log \frac{a}{b}$\n- $\log a + \log b = \log (a \times b)$\nThese rules allow us to combine and simplify logarithmic expressions.\n\n3. **Apply the rules step-by-step:**\nStart with $\log 64 - \log 6 = \log \frac{64}{6}$.\nThen subtract $\log 8$: $\log \frac{64}{6} - \log 8 = \log \frac{64/6}{8} = \log \frac{64}{6 \times 8} = \log \frac{64}{48}$.\nFinally, add $\log \frac{3}{4}$: $\log \frac{64}{48} + \log \frac{3}{4} = \log \left( \frac{64}{48} \times \frac{3}{4} \right)$.\n\n4. **Simplify the fraction inside the logarithm:**\nCalculate $\frac{64}{48} = \frac{4}{3}$ (dividing numerator and denominator by 16).\nThen multiply: $\frac{4}{3} \times \frac{3}{4} = 1$.\n\n5. **Final simplification:**\n$\log 1 = 0$ because the logarithm of 1 in any base is 0.\n\n**Answer:** $0$