Question

How do you solve `8 log_2 2-2log_2 8`?

Answer 1

`8log_2 2 - 2 log_2 8 = 2`

Explanation:

Evaluating `log_2 2` is equivalent to asking
`color(white)("XXX")`"For what value `p=log_2 2` is `2^p = 2`?"
The obvious answer is `log_2 2 = p = 1`

Similarly, evaluating `log_2 8` is equivalent to asking
`color(white)("XXX")`"For what value `q=log_2 8` is `2^q=8`?
with an obvious answer `log_2 8 = q = 3`

So
`color(white)("XXX")8log_2 2 - 2log_2 8`
`color(white)("XXXXXX")=8xx1 - 2xx3`
`color(white)("XXXXXX")=2`