Question

Preciso saber 2^2x-9.2^x+8=0

Answer 1

Boa noite!

Solução!

Para resolver a equação exponencial,vamos usar um artificio.

$2^{2x}-9.2^{x} +8=0\\\\\ (2^{x}) ^{2} -9.2^{x} +8=0\\\\\\ Fazendo!\\\\\\ 2^{x}=y\\\\\\\ (2^{x}) ^{2} -9.2^{x} +8=0\\\\\ (y) ^{2} -9.y+8=0\\\\\\ y ^{2} -9y+8=0$


$y= \dfrac{ 9\pm \sqrt{(-9)^{2}-4.1.8 }}{2.1} \\\\\\\\\ y= \dfrac{ 9\pm \sqrt{81-32 }}{2} \\\\\\\\\ y= \dfrac{ 9\pm \sqrt{49 }}{2} \\\\\\\\\ y= \dfrac{ 9\pm7}{2} \\\\\\\\\ y_{1}=\dfrac{ 9+7}{2}= \dfrac{16}{2}=8\\\\\\\\\ y_{2}=\dfrac{ 9-7}{2}= \dfrac{2}{2}=1$

Retomando o artificio usado na equação,vamos determinar sua solução!

$2^{x}=y\\\\\\ 2^{x}=8\\\\\\ 2^{x}=2^{3} \\\\\\ \boxed{x=3}\\\\\\\ 2^{x}=y\\\\\\\ 2^{x}=2^{0} \\\\\\ \boxed{x=0}$


$\boxed{Resposta:~~S=\{0,3\}}$

Boa noite!
Bons estudos!

Answer 2

2^2x-9.2^x + 8 = 0

2^x = a

a^2 - 9a + 8 = 0

Δ = (-9)² - 4.1.8 = 81 - 32 = 49                             V49 = 7


a = 9+/-7 ===a1 = 8     ;   a2 = 1
        2
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2^x = a1 ==> 2^x1 = 2^3 ==> x1 = 3

2^x = a2 ==> 2^x2 = 2^0 ==> x2 = 0