• Matéria: Matemática
  • Autor: aninharovarisbts
  • Perguntado 5 anos atrás

5^√2^x. 3^√4^x = √8^-x​

Respostas

respondido por: 011bruna
0

Resposta:

Explicação passo-a-passo:

$$\begin{gathered}\sqrt[5]{2^x}.\sqrt[3]{4^x}=\sqrt{8^{-x}}\\\\\mathsf{Como~se~trata~de~pot\^encias~de~multiplo~2,~basta~igualar~as~bases}\\\\2^{\frac{x}{5}}.(2^2)^{\frac{x}{3}}=(2^3)^{\frac{-x}{2}}\\2^{\frac{x}{5}}.2^\frac{2x}{3}=2^\frac{-3x}{2}\\2^{\frac{x}{5}+\frac{2x}{3}}=2^\frac{-3x}{2}\\2^\frac{3x+10x}{15}=2^\frac{-3x}{2}\\\\\frac{13x}{15}=\frac{-3x}{2}\\\\\frac{26x}{\not{30}}=\frac{-45x}{\not{30}}\\\\26x=-45x\\71x=0\\\\S=(0)\end{gathered}$

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