Question

Quanto é 8^x ao quadrado-1=4^x-1

Answer 1

a = 3

b = -2

c = -1

Δ = b² - 4ac

Δ = (-2)² - 4 * 3 * (-1)

Δ = 4 + 12

Δ = 16

x' = (-b + √Δ) / 2a

x' = (-(-2) + √16) / (2 * 3)

x' = (2 + 4) / 6

x' = 6 / 6

x' = 1

x'' = (-b - √Δ) / 2a

x'' = (-(-2) - √16) / (2 * 3)

x'' = (2 - 4) / 6

x'' = (-2) / 6

x'' = -1/3

Portanto, a solução é "x = 1" ou "x = -1/3".

Answer 2

Resposta:

x= -1/3 ou x=1

Explicação passo-a-passo:

$8^{(x^{2}-1)}=4^{(x-1)}\\8=2.2.2=2^{3}\\4=2.2=2^{2}\\\\2^{3(x^{2}-1)}=2^{2(x-1)}\\3(x^{2}-1)=2(x-1)\\3x^{2}-3=2x-2\\3x^{2}-2x-1=0$

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~3x^{2}-2x-1=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=3{;}~b=-2~e~c=-1\\\\\Delta=(b)^{2}-4(a)(c)=(-2)^{2}-4(3)(-1)=4-(-12)=16\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-2)-\sqrt{16}}{2(3)}=\frac{2-4}{6}=\frac{-2\div2}{6\div2}=-\frac{1}{3}\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-2)+\sqrt{16}}{2(3)}=\frac{2+4}{6}=\frac{6}{6}=1$