Question

PLEASE HELP ASAP!!!!

Answer 1

2

Explanation:

Using the rule of logarithms

`log_(b)` x = n ⇒ x =
`b^(n)`

let

`log_(2)` 8 = n ⇒ 8 =
`2^(n)` , that is

2³ =
`2^(n)` ( equate exponents )

n = 3

and let

`log_(3)` (
`(1)/(3)` ) = n ⇒
`(1)/(3)` =
`3^(n)` , that is

`3^(-1)` =
`3^(n)` ( equate exponents )

n = - 1

then

`log_(2)` 8 +
`log_(3)` (
`(1)/(3)` )

= 3 + (- 1)

= 3 - 1

= 2