Решение:
\[ \log_6 2 + \log_6 3 + 2\log_6 4 = \log_6 (2 \times 3) + \log_6 (4^2) \]\[ = \log_6 6 + \log_6 16 \]\[ = 1 + \log_6 16 \]\[ 1 + \log_6 16 \approx 1 + 1.5 = 2.5 \]\[ \log_b \frac{49}{b} = 2.5 \]\[ \log_b 49 - \log_b b = 2.5 \]\[ \log_b 49 - 1 = 2.5 \]\[ \log_b 49 = 3.5 \]\[ b^{3.5} = 49 \]\[ b^{\frac{7}{2}} = 7^2 \]\[ (b^{\frac{7}{2}})^{\frac{2}{7}} = (7^2)^{\frac{2}{7}} \]\[ b = 7^{\frac{4}{7}} \]Ответ: а) 1 + log6 16; б) 7^(4/7)