Question

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

Answer 1

`=1.1402`

Explanation:

Use the definition `log(a/b)=loga-logb`
`log₃(7/2)=log₃7-log₃2=1.7712-0.6310=1.1402`