Question

ALGUEM PODE ME AJUDAR? POR FAVOR?
$log2​​(3x−1)+log2​​(x)=2$
resolve pfvrrrr nao preciso de explicaçao​

Answer 1

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$\Huge\green{\boxed{\rm~~~\blue{S = \{-1, 1.\overline{3}\}~~~}}}$

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$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O\ PASSO{-}A{-}PASSO\ \ \ }}$✍

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☺lá novamente, Iara. Vamos a mais ume exercício❗ Acompanhe a resolução abaixo. Acompanhe a resolução abaixo. ✌

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

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$\LARGE\gray{\boxed{\sf\blue{~~log_2(3x - 1) + log_2(x) = 2~~}}}$

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☔ Temos como propriedade da Função Logaritmo que um logaritmando composto por um produto é igual a soma de logs separados

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$\large\red{\boxed{\pink{\boxed{\begin{array}{rcl}&&\\&\orange{\rm log_c(a \cdot b) \iff log_c(a) + log_c(b) }&\\&&\\\end{array}}}}}$

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$\LARGE\blue{\sf log_2((3x - 1) \cdot x) = 2 }$

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$\LARGE\blue{\sf log_2(3x^2 - x) = 2 }$

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☔ Relembrando que:

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$\large\red{\boxed{\pink{\boxed{\begin{array}{rcl}&&\\&\orange{\rm log_b(a) = c \iff b^c = a}&\\&&\\\end{array}}}}}$

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$\LARGE\blue{\sf 2^2 = 3x^2 - x }$

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$\LARGE\blue{\sf 3x^2 - x - 4 = 0}$

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☔ Pela Fórmula de Bháskara temos:

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$\Large\gray{\boxed{\blue{\sf \pink{\overbrace{3}^{a}}x^2 + \green{\overbrace{(-1)}^{b}}x + \gray{\overbrace{(-4)}^{c}} = 0}}}$

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$\Large\blue{\sf \Delta = (-1)^2 - 4 \cdot 3 \cdot (-4) = 49}$

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$\begin{cases}\large\blue{\text{ \sf x_{1} = \dfrac{-(-1) + \sqrt{49}}{2 \cdot 3} = \dfrac{1 + 7}{6} = 1.\overline{3} }}\\\\\\\large\blue{\text{ \sf x_{2} = \dfrac{-(-1) - \sqrt{49}}{2 \cdot 3} = \dfrac{1 - 7}{6} = -1 }}\end{cases}$

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$\Huge\green{\boxed{\rm~~~\blue{S = \{-1, 1.\overline{3}\}~~~}}}$ ✅

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$\bf\large\red{\underline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

☕ $\bf\Large\blue{Bons\ estudos.}$

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

❄☃ $\sf(\gray{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

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$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$