Question

preciso de ajuda nessa questões.


$9 {}^{x + 1} = \sqrt[3]{3}$
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Answer 1

$\large\boxed{\begin{array}{l}\rm 9^{x+1}=\sqrt[\rm3]{\rm 3}\\\rm (3^2)^{x+1}=3^{\frac{1}{3}}\\\\\rm 3^{2x+2}=3^{\frac{1}{3}}\\\rm 2x+2=\dfrac{1}{3}\\\rm 3\cdot(2x+2)=1\\\rm 6x+6=1\\\rm 6x=1-6\\\rm 6x=-5\\\rm x=-\dfrac{5}{6}\end{array}}$

Answer 2

Explicação passo-a-passo:

$9 {}^{x + 1} = \sqrt[3]{3}$

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

$2x + 2 = \dfrac{1}{3}$

$2x = \dfrac{1}{3} - 2$

$2x = \dfrac{1 - 6}{3}$

$2x = - \dfrac{5}{3}$

$2x \div 2 = - \dfrac{5}{3} \div 2$

$x = - \dfrac{5}{3} \times \dfrac{1}{2}$

$x = - \dfrac{5}{6}$