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(MACKENZIE) Sabendo-se que as raízes da equação logarítmica,
\( \huge\boxed{\boxed{log_{3x} \dfrac{3}{x}+(log_3x)^2=1}}.\\. \)
formam a matriz
\( A= \left|\begin{array}{ccc}x’& x’’’\\18x’’&-1\end{array}\right|. \), então seu
determinante vale:
\( a)~-8\\b)~-6\\c)~5\\d)~-2\\e)~0 \)


RESOLVENDO

1º Passo: Simplificar a equação

\( log_{3x}\frac{3}{x}+(log_{3}x)^{2}=1\\(log_{3}\frac{3}{x})/(log_{3}3x) + (log_{3}x)^{2}=1\\(log_{3}3-log_{3}x)/(log_{3}3+log_{3}x) + (log_{3}x)^{2}=1 \)
2º Passo: Substituir \( log_{3}x \)por \( y \)
\( (1-y)/(1+y)+y^{2}=1\\(1-y)/(1+y)=1-y^{2}\\1-y=(1-y^{2})(1+y)\\1-y=1+y-y^{2}-y^{3}\\0=2y-y^{2}-y^{3}\\y^{3}+y^{2}-2y=0 \)
3º Passo: Encontrar as raízes e substituí-las em \( log_{3}x=y \)
\( y(y^{2}+y-2)=0\\y_{1}=0\\y^{2}+y-2=0\\y_{2}=1\\y_{3}=-2\\log_{3}x_{1}=0\\x_{1}=1\\log_{3}x_{2}=1\\x_{2}=3\\log_{3}x_{3}=-2\\x_{3}=3^{-2}\\x_{3}= \frac{1}{9} \)
4º Passo: Substituir as raízes na matriz e achar o Det(A)

\( A=\left[\begin{array}{cc}x_{1}& x_{3}\\18x_{2}&-1\end{array}\right]\\ A=\left[\begin{array}{cc}1&1/9\\54&-1\end{array}\right]\\ Det|A|=+(1)(-1)-(1/9)(54)\\Det|A| = -1-6\\Det|A| = -7 \)

R: -7(letra A)

\( log_{(3x)}(\frac{3}{x})+(log_{3}x)^{2}=1 \)
Mudando a base do primeiro logaritmo para 3:
\( \dfrac{log_{3}(\frac{3}{x})}{log_{3}(3x)}+(log_{3}x)^{2}=1\\dfrac{log_{3}3-log_{3}x}{log_{3}3+log_{3}x}+(log_{3}x)^{2}=1\\dfrac{1-log_{3}x}{1+log_{3}x}+(log_{3}x)^{2}=1 \)
Multiplicando todos os membros por 1 + log₃x:
\( \dfrac{(1+log_{3}x)(1-log_{3}x)}{1+log_{3}x}+(log_{3}x)^{2}(1+log_{3}x)=1+log_{3}x\\1-log_{3}x+(log_{3}x)^{2}+(log_{3}x)^{3}=1+log_{3}x\\(log_{3}x)^{3}+(log_{3}x)^{2}-log_{3}x-log_{3}x=0\\(log_{3}x)^{3}+(log_{3}x)^{2}-2log_{3}x=0 \)
Colocando log₃x em evidência:
\( log_{3}x\cdot\left[(log_{x})^{2}+log_{3}x-2\right]=0 \)
Logo:
\( log_{3}x=0~~~\therefore~~~3^{0}=x~~~\therefore~~~\boxed{x=1}\\(log_{3}x)^{2}+log_{3}x-2=0\\S=-b/a=-1/1=-1\\P=c/a=-2/2=-2\\log_{3}x’’=1~~~\therefore~~~3^{1}=x’~~~\therefore ~~~\boxed{x’=3}\\log_{3}x’’’=-2~~~\therefore~~~3^{-2}=x’’’’~~~\therefore~~~\boxed{x’’’=\frac{1}{9}} \)
______________________
\( \left|\begin{array}{cc}x’~~~~~~x’’’\\18x’’~~-1\end{array}\right|=\left|\begin{array}{cc}1~~~~~~~~~~\frac{1}{9}\\18\cdot3~~-1\end{array}\right|\\ \left|\begin{array}{cc}x’~~~~~~x’’’\\18x’’~~-1\end{array}\right|=1(-1)-\dfrac{1}{9}\cdot54\\ \left|\begin{array}{cc}x’~~~~~~x’’’\\18x’’~~-1\end{array}\right|=-1-6\\ \left|\begin{array}{cc}x’~~~~~~x’’’\\18x’’~~-1\end{array}\right|=-7 \)



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