Question

How do you approximate `log_5 9` given `log_5 2=0.4307` and `log_5 3=0.6826`?

Answer 1

`log_5 9 approx 1.3652`

Explanation:

We know that

`log_5 9 = log_5 3^2 =2log_5 3 = 2xx0.6826 = 1.3652`