Question

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Determine em R o conjunto solução da equação log3 (x – 2) – log9 (x – 4) = 1. ​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{log_3\:(x - 2) - log_9\:(x - 4) = 1}$

$\mathsf{log_3\:(x - 2) - \dfrac{log_3\:(x - 4)}{log_3\:9} = 1}$

$\mathsf{log_3\:(x - 2) - \dfrac{log_3\:(x - 4)}{log_3\:9} = log_3\:3}$

$\mathsf{log_3\:(x - 2) - \dfrac{log_3\:(x - 4)}{2} = log_3\:3}$

$\mathsf{2\:log_3\:(x - 2) - log_3\:(x - 4)= 2\:log_3\:3}$

$\mathsf{log_3\:(x - 2)^2 - log_3\:(x - 4)= log_3\:3^2}$

$\mathsf{log_3\dfrac{(x - 2)^2}{(x - 4)}= log_3\:9}$

$\mathsf{\dfrac{(x - 2)^2}{(x - 4)}= 9}$

$\mathsf{x^2 - 4x + 4 = 9x - 36}$

$\mathsf{x^2 - 13x + 40 = 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = 13^2 - 4.1.40}$

$\mathsf{\Delta = 169 - 160}$

$\mathsf{\Delta = 9}$

$\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{13 \pm \sqrt{9}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{13 + 3}{2} = \dfrac{16}{2} = 8}\\\\\mathsf{x'' = \dfrac{13 - 3}{2} = \dfrac{10}{2} = 5}\end{cases}}$

$\boxed{\boxed{\mathsf{S = \{8;5\}}}}$