Cho \(\log _{9} 5=a ; \log _{4} 7=b ; \log _{2} 3=c\).Tính \(A=m+2 n+3 p+4 q\)
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Lời giải:
Báo sai\(\begin{aligned} &\text { Ta có } \log _{24} 175=\log _{24} 7.5^{2}=\log _{24} 7+2 \log _{24} 5^{2}=\frac{1}{\log _{7} 24}+\frac{2}{\log _{5} 24} \\ &=\frac{1}{\log _{7} 3+\log _{7} 2^{3}}+\frac{2}{\log _{5} 3+\log _{5} 2^{3}}=\frac{1}{\frac{1}{\log _{3} 7}+\frac{3}{\log _{2} 7}}+\frac{2}{\frac{1}{\log _{3} 5}+\frac{3}{\log _{2} 5}} \\ &=\frac{1}{\frac{1}{\log _{2} 7 \cdot \log _{3} 2}+\frac{3}{\log _{2} 7}}+\frac{2}{\frac{1}{\log _{3} 5}+\frac{3}{\log _{2} 3 \cdot \log _{3} 5}}=\frac{1}{\frac{1}{2 b \cdot \frac{1}{c}}+\frac{3}{2 b}}+\frac{2}{\frac{1}{2 \mathrm{a}}+\frac{3}{\mathrm{c} .2 \mathrm{a}}} \\ &=\frac{1}{\frac{c}{2 b}+\frac{3}{2 b}}+\frac{2}{\frac{c}{2 \mathrm{ac}}+\frac{3}{2 \mathrm{ac}}}=\frac{2 b}{c+3}+\frac{4 \mathrm{ac}}{c+3}=\frac{2 b+4 \mathrm{a} c}{c+3} . \\ &A=m+2 n+3 p+4 q=2+8+3+12=25 \end{aligned}\)