Question

DETERMINE NO CAMPO DOS NÚMEROS REAIS, A EQUAÇÃO



$\large\boxed{\log_x(3x)+\log_9(x)=1}$

Answer 1

 Passando para a base 10 ficamos com:

$\log_x (3x) + \log_9 x = 1 \\\\\\ \frac{\log 3x}{\log x} + \frac{\log x}{\log 9} = 1 \\\\\\ \frac{\log 3 + \log x}{\log x} + \frac{\log x}{2 \cdot \log 3} = 1$

 Fazendo $\log 3 = \delta$ e $\log x = \rho$, temos que:

$\frac{\delta + \rho}{\rho} + \frac{\rho}{2\delta} = 1 \\\\\\ (\delta + \rho) \cdot 2\delta + \rho^2 = 2 \delta \cdot \rho \\\\ \rho^2 = - 2 \cdot \delta^2 \\\\ \frac{\rho^2}{\delta^2} = - 2$

 Como podemos notar, não existe $\left \{\delta, \; \rho \; \in \mathbb{R} | \frac{\rho^2}{\delta^2} = - 2{ \right\}$

 Daí, $\boxed{S=\left\{\right\}}$.

Answer 2

Explicação passo-a-passo:

$\sf log_{x}~(3x)+log_{9}~(x)=1$

$\sf \dfrac{log_{3}~(3x)}{log_{3}~(x)}+\dfrac{log_{3}~(x)}{log_{3}~(9)}=1$

$\sf \dfrac{log_{3}~(3)+log_{3}~(x)}{log_{3}~(x)}+\dfrac{log_{3}~(x)}{2}=1$

$\sf \dfrac{1+log_{3}~(x)}{log_{3}~(x)}+\dfrac{log_{3}~(x)}{2}=1$

Seja $\sf y=log_{3}~(x)$

$\sf \dfrac{1+y}{y}+\dfrac{y}{2}=1$

$\sf 2\cdot(1+y)+y^2=2y$

$\sf 2+2y+y^2=2y$

$\sf y^2=-2$

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Logo, $\sf S=\{~\}$