Question

${2}^{x - 1} = \sqrt2{ \sqrt[3]{2} } \div 8 \times \sqrt[4]{2}$
x é igual a:

(a) 3
(b) -7/12
(c) -43/12
(d) 5/4
(e) -19/12​

Answer 1

Explicação passo-a-passo:

${2}^{x - 1} = \sqrt{2}. \sqrt[3]{2} \div 8 \times \sqrt[4]{2} => {2}^{x - 1} = \sqrt[6]{4} \div {2}^{3} \times \sqrt[4]{2} = > {2}^{x - 1} = ( {2}^{2})^{ \frac{1}{6} } \div {2}^{3} \times {2}^{ \frac{1}{4} } = > {2}^{x - 1} = {2}^{( \frac{2}{6} - 3 + \frac{1}{4})} = > {2}^{x - 1} = {2}^{( \frac{4 - 36 + 3}{12}) } = > {2}^{x - 1} = {2}^{ \frac{ - 29}{12} } = > x - 1 = \frac{ - 29}{12} = > x = \frac{ - 29}{12} + 1 = > x = \frac{ - 29 + 12}{12} = > x = \frac{ - 17}{12}$.