`\log_{9}5=a;\log_{2}7=b;\log_{4}12=c ` Tính `\log_{18}4200` theo `a,b,c`

Ta có: `log_4 12=c`

`=>` `log_2 12=2c`

`=>` `log_2 (4*3)=2c`

`=>` `2+log_2 3=2c`

`=>` `log_2 3=2c-2`

`=>` `log_3 2=1/(2c-2)`

`log_18 4200`

`=log_18 (7*12*2*5^2)`

`=log_18 7+log_18 12 + log_18 2 + 2log_18 5`

`=(log_2 7)/(log_2 18)+(log_4 12)/(log_4 18) + 2*(log_9 5)/(log_9 18)+1/(log_2 18)`

`=(b)/(log_2 (2*3^2))+(c)/(1/2log_2 (2*3^2))+2*(a)/(log_9 (2*3^2))+1/(log_2 (2*3^2))`

`=(b)/(1+2log_2 3)+(2c)/(1+2log_2 3)+(4a)/(2+log_3 2)+1/(1+2log_2 3)`

`=b/(1+2(2c-2))+(2c)/(1+2(2c-2))+(4a)/(2+1/(2c-2))+1/(1+2(2c-2))`

`=(b+2c+8ac-8a+1)/(4c-3)`