Question

(PUCRS) Se 27 elevado na log de x na base 9 é igual a 1/2, então x é igual a ?

Answer 1

$\displaystyle 27^{\log_9x}=\frac{1}2\\ \\ \log_327^{\log_9x}=\log_3\frac{1}2\\ \\ \log_9x\cdot\log_327=\log_3\frac{1}2\\ \\ 3\log_9x=\log_3\frac{1}2\\ \\ \frac{3}{2}\log_3x=\log_3\frac{1}2\\ \\ \frac{3}{2}\log_3x=\log_3\frac{1}2$

$\displaystyle \log_3x=\log_3\left(\frac{1}2\right)^{2/3}\\ \\ \\ \boxed{x=\sqrt[3]{\frac{1}{4}}}$

Answer 2

Boa Tarde,

vamos usar as propriedades, da potência, da mudança de base e da definição,

$log(b)^a\Rightarrow a\cdot log(b)\\\\ log_b(a)\Rightarrow \dfrac{log(a)}{log(b)}\\\\ log_b(c)=a\Rightarrow c=b^a$

_____________


$27^{\log_9(x)}= \dfrac{1}{2}\\\\ \log_3(27^{\log_9(x)})=\log_3\left( \dfrac{1}{2}\right)\\\\ \log_3(27)^{ \tfrac{\log_3(x)}{\log_39}}=\log_3\left( \dfrac{1}{2}\right)\\\\ \left[ \dfrac{\log_3(x)}{\log_39}\right]\cdot\log_327=\log_3\left( \dfrac{1}{2}\right)\\\\ 3\cdot\left( \dfrac{\log_3(x)}{2}\right)=\log_3\left( \dfrac{1}{2}\right)\\\\ \dfrac{3\log_3(x)}{2}=\log_3\left( \dfrac{1}{2}\right)\\\\ 3\log_3(x)=2\log_3\left( \dfrac{1}{2}\right)$ 

Agora, eliminamos as bases, igualando a equação..

$\log_3(x)= \dfrac{2}{3}\log_3\left( \dfrac{1}{2}\right)\\\\ x=\left( \dfrac{1}{2}\right)^{ \tfrac{2}{3}}\\\\ x=2^{- \tfrac{2}{3}}\\\\ x= \sqrt[3]{2^{-2} } \\\\ x= \sqrt[3]{ \dfrac{1}{4} }\\\\ x= \dfrac{ \sqrt[3]{1} }{ \sqrt[3]{4} }= \dfrac{1\cdot \sqrt[3]{4} }{ \sqrt[3]{4}\cdot \sqrt[3]{4} }= \dfrac{ \sqrt[3]{4} }{4}$

Tenha ótimos estudos ;D