\(log_{18}4200=\dfrac{log_24200}{log_218}=\dfrac{log_2\left(2^3.3.5^2.7\right)}{log_2\left(2.3^2\right)}=\dfrac{3+log_23+2log_25+log_27}{1+2log_23}\)
\(log_412=log_{2^2}\left(2^2.3\right)=\dfrac{1}{2}log_2\left(2^2.3\right)=\dfrac{1}{2}\left(2+log_23\right)=c\)
\(\Rightarrow log_23=2c-2\)
\(log_25=\dfrac{log_95}{log_92}=log_95.log_29=2log_23.log_95=2a\left(2c-2\right)=4ac-4a\)
Thế vào trên: \(log_{18}4200=\dfrac{3+2c-2+2\left(4ac-4a\right)+b}{1+2\left(2c-2\right)}\)
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