Question

How do you approximate `log_5 8` given `log_5 2=0.4307` and `log_5 3=0.6826`?

Answer 1

I found: `1.2921`

Explanation:

We can write:
`log_5(8)=log_5(2*2*2)=log_5(2)+log_5(2)+log_5(2)=3log_5(2)=3*0.4307=1.2921`