Question

1) Resolva as expressões : A) log x36 = 2 ; B) log 1/2 (x-2) = -3 ; C) log 2 (x2+x+2) = 3

Answer 1

A) Temos que logₓ 36 = 2 => x² = 36 => x = √36 => x = 6, pois a base do logaritmo deve ser positivo e diferente de 1.

B) log$_{1/2}$(x -2) = -3 => $(\frac{1}{2})^{-3}=(x - 2)=>2^{3}=x-2=>8=x-2->x=8+2=>x=10$

C) $log_{2}(x^{2}+x+2)=3=> 2^{3}=x^{2}+x+2=>x^{2}+x+2=8=> x^{2}+x-6=0$, onde

a = 1, b = 1 e c = -6

Δ = 1² - 4.1.(-6)

Δ = 1 + 24

Δ = 25

x = (-1 +ou - √25)/2.1

x' = (-1 + 5)/2 => x' = 4/2 => x' = 2

x" = (-1 -5)/2 => x" = -6/2 => x" = -3