Question

Encontre o valor de X que torna as igualdades verdadeiras:

Answer 1

$\sqrt[5]{10^3} = \sqrt[x]{10^9} \\ \\ 10^{ \frac{3}{5}}=10^{ \frac{9}{x} } \\ \\ {\not10}^{ \frac{3}{5}}=\not10^{ \frac{9}{x} } \\ \\ \frac{3}{5} = \frac{9}{x} \\ \\ 9x^{-1}= \frac{3}{5} \\ \\ x^-^1= \frac{ \frac{3}{5}}{9} \\ \\ x^-^1= \frac{1}{15} \\ \\ x= \frac{ \frac{1}{1} }{15} \\ \\ x=15$
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$\sqrt[8]{2^{20}} = \sqrt{2^x} \\ \\ 2^{ \frac{20}{8} }=2^{ \frac{x}{2}} \\ \\ \not2^{ \frac{20}{8} }=\not2^{ \frac{x}{2}} \\ \\ \frac{20}{8} = \frac{x}{2} \\ \\ \frac{5}{2} = \frac{1}{2} *x \\ \\ \frac{\frac{1}{2} *x}{\frac{1}{2}}= \frac{\frac{5}{2}}{ \frac{1}{2}} \\ \\ \frac{5}{2} * \frac{2}{1} = \frac{10}{2} \>\>\>\>\>=5$