Question

How do you evaluate `4log_4 6 - log_4 5`?

Answer 1

Use the following properties of logs:
1] `alogx=logx^a`
2] `logx-logy=log(x/y)`
So in your case:
`log_4(6^4)-log_4(5)=log_4(6^4/5)=log_4(259.2)=x`
so:
`4^x=259.2`
if `x=4`
`4^4=256`
if `x=4.002`
`4^(4.002)=259.2`

Another thing that you can do is to change base (using, for example, natural logs) as:
`log_4(259.2)=ln(259.2)/(ln(4))=4.0089`