Question

If $\log 4=0.602$ and $\log 27=1.431$ then the value of $\log 8$ is $k$,then

last digit of decimal of $k$ is

Answer 1

$\log 4=0.602\log 2^2=0.6022\log 2=0.602\log 2=0.301\log 8=\log 2^3=3\log 2=3\times 0.301=0.903$