Question

39. DETERMINE
valor de X em cada caso
$\sqrt{7} = \sqrt[8]{x}$
$\sqrt[12]{36} = \sqrt[x]{6}$
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Answer 1

Resposta:

Explicação passo-a-passo:

1)

$\sqrt{7} = \sqrt[8]{x} \\(\sqrt{7})^{8} =(\sqrt[8]{x} )^{8} \\logo \\x= (\sqrt{7})^{8}$

2)

$\sqrt[12]{36}=\sqrt[x]{6} \\\sqrt[12]{6.6} = 6^{\frac{1}{x} } \\(6.6)^{\frac{1}{12} } = 6^{\frac{1}{x} } \\\\6^{\frac{1}{12} } .6^{\frac{1}{12} } =6^{\frac{1}{x} }\\6^{\frac{1}{12} } =\frac{ 6^{\frac{1}{x} } }{6^{\frac{1}{12} } } \\\\6^{\frac{1}{12} } =6^{\frac{1}{x} - \frac{1}{12} }$

bases iguais iguala os expoentes.

$\frac{1}{12} = \frac{1}{x} - \frac{1}{12} \\\frac{1}{12} + \frac{1}{12} =\frac{1}{x} \\\frac{2}{12} = \frac{1}{x} \\2x=12\\x=\frac{12}{2} \\x= 6$