Question

resolva a equação 2 elevevado a 3x-1=32elevado a 2x

Answer 1

$2 {}^{3x - 1} = 32 {}^{2x} \\ \\ 2 {}^{3x - 1} =( 2 {}^{5} ) {}^{2x} \\ \\ {2}^{3x - 1} = {2}^{10x} \\ \\ \\ \\ 3x - 1 = 10x \\ \\ 3x - 1 - 10x + 1 = 10x - 10x + 1 \\ \\ - 7x = 1 \\ \\ \frac{ - 7x}{ - 7} = \frac{1}{ - 7} \\ \\ x = - \frac{1}{7} \\ \\ \\ \\ \\ \\ \\ \\ \\ 2 {}^{3 \times ( - \frac{ 1}{7} ) - 1} = 2 {}^{10 \times ( - \frac{1}{7} )} \\ \\ 2 {}^{ - \frac{3}{7} - \frac{7}{7} } = 2 {}^{ - \frac{10}{7} } \\ \\ 2 {}^{ - \frac{10}{7} } = 2 {}^{ - \frac{10}{7} } \\ \\ ( \frac{1}{2} ) {}^{ \frac{10}{7} } = ( \frac{1}{2} ) {}^{ \frac{10}{7} } \\ \\ \sqrt[7]{( \frac{1}{2}) {}^{10} } =\sqrt[7]{( \frac{1}{2}) {}^{10} } \\ \\ \frac{1}{2} \sqrt[7]{( \frac{1}{2}) {}^{3} } = \frac{1}{2} \sqrt[7]{( \frac{1}{2}) {}^{3} } \\ \\ \frac{1}{2} \sqrt[7]{ \frac{1}{8} } = \frac{1}{2} \sqrt[7]{ \frac{1}{8} }$

$x = - \frac{ 1}{7}$