Question

A solução da equação 4^x+1=2^3-2x é??

Answer 1

Boa Tarde,


acompanhe passo a passo a resolução, aplicando as propriedades da exponenciação:

$\Large\underbrace{\mathsf{4^{x+1}=2^{3-2x}}}\\\\ \mathsf{(2^2)^{x+1}=2^3\cdot2^{-2x}}\\\\ \mathsf{2^{2x+2}=8\cdot \dfrac{1}{2^{2x}} }\\\\ \mathsf{2^{2x}\cdot2^2=\dfrac{8}{2^{2x}} }\\\\ \mathsf{4\cdot2^{2x}= \dfrac{8}{2^{2x}} }\\\\\\ \mathsf{2^{2x}=y}\\\\\\ \mathsf{4\cdot y= \dfrac{8}{y} }\\\\ \mathsf{4y= \dfrac{8}{y} }\\\\ \mathsf{y\cdot 4y=8}\\ \mathsf{4y^2=8}\\\\ \mathsf{y^2= \dfrac{8}{4} }\\\\ \mathsf{y^2=2}\\ \mathsf{y=\pm \sqrt{2} }\\\\\\ \mathsf{2^{2x}=y~~Lembra?}$


$\mathsf{2^{2x}= \sqrt{2} }}~~~~~~~~~~~~~~~~~~\mathsf{2^{2x}=- \sqrt{2} }~~(\notin\mathbb{R})}\\\\ \mathsf{2^{2x}= 2^{ \tfrac{1}{2} }}\\\\ \mathsf{\not2^{2x}= \not2^{ \tfrac{1}{2} }}\\\\ \mathsf{{2x}= \dfrac{1}{2} }}\\\\ \mathsf{x= \dfrac{1}{4} }\\\\ \mathsf{Portanto:}\\\\\\ \Large\boxed{\mathsf{S=\left\{ \dfrac{1}{4} \right\}}}$

tenha ótimos estudos ;D