Question

How do you evaluate `log_8 4`?

Answer 1

Consider that:
`log_a(b)=x ->a^x=b`
in your case:
`log_8(4)=x`
So:
`8^x=4` that can be written as:
`(2^3)^x=2^2`
`2^(3x)=2^2`
so that:
`3x=2`
and `x=2/3`