Question

How do you approximate `log_3 (18)` given `log_3 2=0.6310` and `log_3 7=1.7712`?

Answer 1

`log_3(18) ~~ 2.6310`

Explanation:

`log_3(18) = log_3(2*3*3)= log_3(2)+log_3(3)+log_3(3)`

`~~ 0.6310+1+1 = 2.6310`