Question

Em cada caso, determine o valor de x. ​

Answer 1

Resposta:

Explicação passo-a-passo:

A) $\sqrt[3]{2^{15}} = \sqrt[5]{2^{x}} => 2^{\frac{15}3} = 2^{\frac{x}{5}} =>$ $\frac{15}{3} = \frac{x}{5} =>$ 15× 5= 3x=> x= $\frac{75}{3}$=> x= 25

B) $\sqrt[30]{3^{10}} = \sqrt[x]{3} => 3^{\frac{10}{30} } = 3^{\frac{1}{x}} => \frac{10}{30} = \frac{1}{x} => \frac{1}{3} =\frac{1}{x} => x=3$

C)$\sqrt[21]{3^{7}}=\sqrt[6]{3^{x}}=> 3^{\frac{7}{21}}= 3^{\frac{x}{6}}=>\frac{7}{21} = \frac{x}{6}=> \frac{1}{3}=\frac{x}{6}=>x=\frac{6}{3}=> x=2$

D)$\sqrt[24]{13^{8}}=\sqrt[x]{13}=> 13^{\frac{8}{24}}= 13^\frac{1}{x}=> \frac{8}{24}=\frac{1}{x} => \frac{1}{3}= \frac{1}{x}=>x=3$

E)$\sqrt[6]{64}=\sqrt[3]{2^x}=> \sqrt[6]{2^6}=\sqrt[3]{2^x}=> 2^{\frac{6}{6}}= 2^{\frac{3}{x}}=> \frac{6}{6}= \frac{3}{x}=> 1= \frac{3}{x}=> x=3$

F)$\sqrt[9]{64}= \sqrt[x]{4}=> \sqrt[9]{2^{6}}= \sqrt[x]{2^{2}}=>x^{\frac{6}{9}}=x^{\frac{2}{x}}=> \frac{6}{9}= \frac{2}{x}=> \frac{2}{3}= \frac{2}{x}=> \frac{1}{3}=\frac{1}{x}=> x=3$