Question

Se $(0,4)^{4x+1} = \sqrt[3]{ \frac{5}{2} }$ , então "x" vale:     a) -$\frac{1}{6}$      b) -$\frac{1}{3}$    c) -$\frac{1}{2}$    d) $\frac{1}{2}$   e)  $\frac{1}{5}$

Answer 1

Olá Thalyra:

$(0,4)^{4x+1} = \sqrt[3]{ \frac{5}{2} }$

$( \frac{4}{10} )^{4x+1} = (\frac{5}{2} )^{ \frac{1}{3} }$


$( \frac{2}{5} )^{4x+1} = (\frac{5}{2} )^{ \frac{1}{3} }$

$( \frac{2}{5} )^{4x+1} = (\frac{2}{5} )^{ \frac{-1}{3} }$

$4x + 1 = \frac{-1}{3}$

$12x + 3 = -1 12x = -4$

$x = \frac{-4}{12}$

$x = \frac{-1}{3}$

Answer 2

$(\frac{2}{5})^{4x+1}= \sqrt[3]{\frac{5}{2}}$

$(\frac{2}{5})^{4x+1}=(\frac{5}{2})^{\frac{1}{3}}$

$\not(\frac{2}{5})^{4x+1}=\not(\frac{2}{5})^{-\frac{1}{3}}$

$4x+1=-\frac{1}{3}\\ \\12x+3=-1\\12x=-4$

$x=\frac{-4^{\div4}}{12^{\div4}}$

$\boxed{x=\frac{-1}{3}}$