Question

(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:

$a)~-8\\b)~-6\\c)~5\\d)~-2\\e)~0$

Answer 1

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)

Answer 2

$log_{(3x)}(\frac{3}{x})+(log_{3}x)^{2}=1$

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:

$log_{3}x=0~~~\therefore~~~3^{0}=x~~~\therefore~~~\boxed{x=1}\\\\\\(log_{3}x)^{2}+log_{3}x-2=0\\S=-b/a=-1/1=-1\\P=c/a=-2/2=-2\\\\log_{3}x''=1~~~\therefore~~~3^{1}=x'~~~\therefore ~~~\boxed{x'=3}\\log_{3}x'''=-2~~~\therefore~~~3^{-2}=x''''~~~\therefore~~~\boxed{x'''=\frac{1}{9}}}$
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$\left|\begin{array}{cc}x'~~~~~~x'''\\18x''~~-1\\\end{array}\right|=\left|\begin{array}{cc}1~~~~~~~~~~\frac{1}{9}\\18\cdot3~~-1\\\end{array}\right|\\\\\\\left|\begin{array}{cc}x'~~~~~~x'''\\18x''~~-1\\\end{array}\right|=1(-1)-\dfrac{1}{9}\cdot54\\\\\\\left|\begin{array}{cc}x'~~~~~~x'''\\18x''~~-1\\\end{array}\right|=-1-6\\\\\\\left|\begin{array}{cc}x'~~~~~~x'''\\18x''~~-1\\\end{array}\right|=-7$