Resposta:

[tex]\textsf{Leia abaixo}[/tex]

Explicação passo a passo:

[tex]\mathsf{4^x - 2^{(x + 3)} = 2^7}[/tex]

[tex]\mathsf{(2^2)^x - 2^x.2^3 = 2^7}[/tex]

[tex]\mathsf{(2^x)^2 - 8.2^x - 128 = 0}[/tex]

[tex]\mathsf{y = 2^x}[/tex]

[tex]\mathsf{y^2 - 8y - 128 = 0}[/tex]

[tex]\mathsf{\Delta = b^2 - 4.a.c}[/tex]

[tex]\mathsf{\Delta = (-8)^2 - 4.1.(-128)}[/tex]

[tex]\mathsf{\Delta = 64 + 512}[/tex]

[tex]\mathsf{\Delta = 576}[/tex]

[tex]\mathsf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{8 \pm \sqrt{576}}{2} \rightarrow \begin{cases}\mathsf{y' = \dfrac{8 + 24}{2} = \dfrac{32}{2} = 16}\\\\\mathsf{y'' = \dfrac{8 - 24}{2} = -\dfrac{16}{2} = -8}\end{cases}}[/tex]

[tex]\mathsf{2^x = 16}[/tex]

[tex]\mathsf{\not2^x = \not2^4}[/tex]

[tex]\mathsf{x = 4}[/tex]

[tex]\boxed{\boxed{\mathsf{S = \{4\}}}}[/tex]

Verified answer

Resposta:

4^x -2^(x+3) = 2^7

(2²)^x -2^(x) * 2³ =2^7

2^(2x) -8*2^(x) =128

(2^x)² -8*2^(x) -128 =0

fazendo y=2^x

y²-8y-128=0

y'=[8+√(64+512)]/2 =(8+24)/2=16=2^4

y''=[8-√(64+512)]/2 =(8-24)/2=-8

y=2^4= 2^x    ==>x=4

y=-8=2^x  ==>não existe x Real  possível

Resposta ==> x=4

#### verificando

4^4 -2^(4+3) = 2^7

256-2^7=128

256-128=128

128 = 128    igualdade verdadeira