Question

PLEASE HELP ASAP!!!!

Answer 1

Answer:

2

Step-by-step 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

$log_{3}$ ( $\frac{1}{3}$ ) = n ⇒ $\frac{1}{3}$ = $3^{n}$ , that is

$3^{-1}$ = $3^{n}$ ( equate exponents )

n = - 1

then

$log_{2}$ 8 + $log_{3}$ ($\frac{1}{3}$ )

= 3 + (- 1)

= 3 - 1

= 2