jika 7log2=a dan 2log3=b maka 6log98=

[tex]
^6\log{98}\\
=\frac{^2\log{98}}{^2\log{6}}\\
=\frac{^2\log{(2\cdot7^2)}}{^2\log{(2\cdot3)}}\\
=\frac{^2\log{2}+2\cdot^2\log{7}}{^2\log{2}+^2\log{3}}\\
=\frac{1+2a}{1+b}\\
[/tex]