Answer:
2
Step-by-step explanation:
Using the rule of logarithms
[tex]log_{b}[/tex] x = n ⇒ x = [tex]b^{n}[/tex]
let
[tex]log_{2}[/tex] 8 = n ⇒ 8 = [tex]2^{n}[/tex] , that is
2³ = [tex]2^{n}[/tex] ( equate exponents )
n = 3
and let
[tex]log_{3}[/tex] ( [tex]\frac{1}{3}[/tex] ) = n ⇒ [tex]\frac{1}{3}[/tex] = [tex]3^{n}[/tex] , that is
[tex]3^{-1}[/tex] = [tex]3^{n}[/tex] ( equate exponents )
n = - 1
then
[tex]log_{2}[/tex] 8 + [tex]log_{3}[/tex] ([tex]\frac{1}{3}[/tex] )
= 3 + (- 1)
= 3 - 1
= 2
