Question

How do you evaluate `log_64 (1/2)`?

Answer 1

`-1/6`

Explanation:

Recall that `log_2(8)=3, => 2^3=8`

let `log_64(1/2)=x`

`64^x=1/2`

`64=2^6`
`64^x=2^(6x)`
`1/2=2^-1`

`2^(6x)=2^-1`
`6x=-1`
`x=-1/6`