Question

Given $\log 2={}^{.}3010300,\log 3={}^{.}4771213,\log 7={}^{.}8450980,$ find the value of
$\log 84$.

Answer 1

$\log 84=\log (4\times 21)=2\log 2+\log \left(21\right)=2\log 2+\log 3+\log 7=2\times \left(0.3010300\right)+0.4771213+0.8450980=1.9242793$