Question

Calcule log2 [logx (x+2)]=1

Answer 1

Propriedade fundamental dos logaritmos:

$\displaystyle \log_{ a }{ c } = b \Rightarrow a^b = c$

Assim:

$\displaystyle \log_{ 2 }[{ \log_{ x }{ x+2 } } ] = 1$

$\displaystyle 2^1 = \log_{ x }{ x+2 }$

$\displaystyle \log_{ x }{ x+2 } = 2$

$\displaystyle x^2 = x+2$

$\displaystyle x^2 - x - 2 = 0$

Por Bhaskara vamos ter:

Δ = (-1)²- 4.1.(-2) = 1 + 8 = 9

x' = - (-1) + √9 / 2.1 = 1 + 3 / 2 = 4/2 = 2

x'' = 1 - 3 / 2 = -2/2 = -1

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