Question

If $\log 4=0.602$ and $\log 27=1.431$, then the value of $\log 12$ is $1.0k9$ where $k$ is 2nd digit in decimalexpression ,then $k$is

Answer 1

$\log 4=0.602\log 27=\log 3^3=1.4313\log 3=1.431\log 3=1.431/3=0.477\log 12=\log (4\times 3)=\log 4+\log 3=0.602+0.477=1.079$