Cho log 9 5 = a; log 4 7 = b; log 2 3 = c. Biết log 24 175 = (mb + nac)/(pc + q. Tính A = m + 2n + 3p + 4q

Ta có \[{\log _{24}}175 = {\log _{24}}\left( {7 \cdot {5^2}} \right) = {\log _{24}}7 + 2{\log _{24}}5 = \frac{1}{{{{\log }_7}24}} + \frac{2}{{{{\log }_5}24}}\]

\[ = \frac{1}{{{{\log }_7}3 + {{\log }_7}{2^3}}} + \frac{2}{{{{\log }_5}3 + {{\log }_5}{2^3}}} = \frac{1}{{\frac{1}{{{{\log }_3}7}} + \frac{3}{{{{\log }_2}7}}}} + \frac{2}{{\frac{1}{{{{\log }_3}5}} + \frac{3}{{{{\log }_2}5}}}}\]

\[ = \frac{1}{{\frac{1}{{{{\log }_2}7 \cdot {{\log }_3}2}} + \frac{3}{{{{\log }_2}7}}}} + \frac{2}{{\frac{1}{{{{\log }_3}5}} + \frac{3}{{{{\log }_2}3 \cdot {{\log }_3}5}}}} = \frac{1}{{\frac{1}{{2b \cdot \frac{1}{c}}} + \frac{3}{{2b}}}} + \frac{2}{{\frac{1}{{2a}} + \frac{3}{{c \cdot 2a}}}}\]

\[ = \frac{1}{{\frac{c}{{2b}} + \frac{3}{{2b}}}} + \frac{2}{{\frac{c}{{2ac}} + \frac{3}{{2ac}}}} = \frac{{2b}}{{c + 3}} + \frac{{4ac}}{{c + 3}} = \frac{{2b + 4ac}}{{c + 3}}\].

Suy ra \(m = 2,n = 4,p = 1,q = 3\). Vậy \[A = m + 2n + 3p + 4q = 2 + 8 + 3 + 12 = 25\].

Đáp án: 25.