Question

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Answer 1

Respuesta:

Resuelve

$9=\sqrt{\frac{3^x}{\sqrt{3}}}$

$2=\left(x-\frac{1}{2}\right)\frac{1}{2}$

$x=\frac{9}{2}$

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Hallar  "x"  

$\sqrt{8}^{x+3}=\sqrt[4]{32}^{3x+1}$

$\frac{3}{2}\left(x+3\right)=\frac{5}{4}\left(3x+1\right)$

$x=\frac{13}{9}$

Saludos...