Question

A solução da equação :

$\sqrt[3]{2 ^{(2.x+5)} } =(0,25) ^{-2.x} \ ?$

Answer 1

$\sqrt[3]{2^{2x+5}}=(0.25)^{-2x}\\\\2^{\frac{2x+5}{3}}=(\frac{1}{4})^{-2x}\\\\2^{\frac{2x+5}{3}}=(2^{-2})^{-2x}\\\\2^{\frac{2x+5}{3}}=2^{4x}\\\\\frac{2x+5}{3}=4x\\\\2x+5=12x\\\\10x=5\\\\\boxed{x=\frac{1}{2}}$

Answer 2

vamos lá!

$\sqrt[3]{2 ^{(2.x+5)} } =(0,25) ^{-2.x} \ \\ \\ {2}^{ \frac{2x + 5}{3} } = ( {2})^{( - 1).( - 2x) } \\ \\ {2}^{ \frac{2x + 5}{3} } = {2}^{2x} \\ \\ \frac{2x + 5}{3} = 2x \\ \\ 2x + 5 = 3.(2x) \\ \\ 2x + 5 = 6x \\ \\ 2x - 6x = - 5 \\ \\ - 4x = - 5 \\ \\ x = \frac{ - 5}{ - 4} \\ \\ x = \frac{5}{4}$

espero ter ajudado!

boa noite!