Question

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

Answer 1

Propriedades usadas:

$log_{(b^{n})}(a)=\dfrac{1}{n}log_{b}(a)\\\\\\x\cdot log_{b}(a)\rightleftharpoons log_{b}(a^{x})\\\\\\a^{log_{a}(b)}=b$
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$27^{log_{9}(x)}=\dfrac{1}{2}\\\\\\(3^{3})^{log_{(3^{2})}(x)}=\dfrac{1}{2}\\\\\\(3^{3})^{\frac{1}{2}log_{3}(x)}=\dfrac{1}{2}\\\\\\3^{\frac{3}{2}log_{3}(x)}=\dfrac{1}{2}\\\\\\3^{log_{3}(x^{3/2})}=\dfrac{1}{2}$

Utilizando a propriedade $a^{log_{a}(b)}=b$:

$x^{\frac{3}{2}}=\dfrac{1}{2}$

Elevando os dois lados da equação a 2/3:

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

Racionalizando a resposta:

$x=\dfrac{1}{\sqrt[3]{4}}$

Multiplicando o numerador e o denominador por (∛4)²:

$x=\dfrac{1\cdot\sqrt[3]{4^{2}}}{\sqrt[3]{4}\cdot(\sqrt[3]{4})^{2}}\\\\\\x=\dfrac{\sqrt[3]{2^{4}}}{(\sqrt[3]{4})^{3}}\\\\\\x=\dfrac{\sqrt[3]{2^{3}\cdot2^{1}}}{4}\\\\\\x=\dfrac{\sqrt[3]{2^{3}}\cdot\sqrt[3]{2^{1}}}{4}\\\\\\x=\dfrac{2\sqrt[3]{2}}{4}\\\\\\\boxed{\boxed{x=\dfrac{\sqrt[3]{2}}{2}}}$