296 4) 3 log 7 21log7 64 2 4 log52 +log5 27 3

$$\frac{3log_7 2 - \frac{1}{2}log_7 64}{4log_5 2 + \frac{1}{3}log_5 27} = \frac{log_7 2^3 - log_7 64^{\frac{1}{2}}}{log_5 2^4 + log_5 27^{\frac{1}{3}}} = \frac{log_7 8 - log_7 8}{log_5 16 + log_5 3} = \frac{0}{log_5 (16 \cdot 3)} = \frac{0}{log_5 48} = 0$$.

Ответ: 0