Question

Como resolvo exercícios desse tipo ?
Consegui resolver vários, mas tem alguns quase impossíveis kkkkkkkkkkk

$( \sqrt[3]{3^x})^x = \sqrt[3]{81} \\ \\ \sqrt{3^(x-1)} = ( \sqrt[3]{9^x} )^2$

Answer 1

$\sqrt[n]{a^m} = a^{ \frac{m}{n} }$

aplicando isso
$( \sqrt[3]{3^x} )^x= \sqrt[3]{81}$

81= 3*3*3*3

$(3^{ \frac{x}{3} })^x= (3^{4})^{ \frac{1}{3} }\\\\ 3^\frac{x^2}{3}= 3^{ \frac{4}{3} } \\\\\\ \frac{x^2}{3}= \frac{4}{3} \\\\x^2=4\\\\\ x =\pm \sqrt{4} \\\\x=\pm 2$

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$\sqrt{3^{x-1}} = ( \sqrt[3]{9^x} )^2\\\\ \sqrt{3^{x-1}} = ( \sqrt[3]{3^{2x}} )^2\\\\ 3^{ \frac{x-1}{2} }= (3^{ \frac{2x}{3} })^2\\\\ 3^{ \frac{x-1}{2} }= 3^\frac{4x}{3} \\\\ \frac{x-1}{2} = \frac{4x}{3} \\\\(x-1)*3 = 4x*2\\\\3x-3 = 8x\\\\-3 =8x-3x\\\\-3 = 5x\\\\ \frac{-3}{5}=x$

Answer 2

$(\sqrt[3]{ x^{x} }) ^x = \sqrt[3]{81}$ 
$\sqrt[3]{3^x^2} = \sqrt[3]{3^4}$

$3^{ \frac{x^2}{3} } = 3^{ \frac{4}{3} }$

x²/3 = 4/3
3x² = 12
 x² = 12/3
x² = 4
 x = √4
 x = 2