2log7=a dan 2log3=b, maka 6log14 adalah...

[tex]^2log7=a , ^2log3=b \\ ^6log14= \frac{^2log14}{^2log6} = \frac{^2log2 +^2log7}{^2log2+^2log3} = \frac{1+a}{1+b} [/tex]