Question

resolva: 27^(x^(2) +1) = 9^(5x)​

Answer 1

Note que :

27 = 3³

9 = 3²

Assim temos :

27^(x² + 1) = 9^(5x)

3^3.(x² + 1) = 3^2.(5x)

Assim temos:

3.(x² + 1) = 2 . 5x

3x² + 3 = 10x

3x² - 10x + 3 = 0

Δ = 100 - 36

Δ = 64

x = (10 +- √64)/6

x = (10 +- 8)/6

x' = 18/6

x' = 3

x'' = 2/6

x'' = 1/3 S = { 1/3 , 3 }

Answer 2

Resposta:

S = {x∈R | x' = 1/3, x'' = 3}

ou

S = {1/3, 3}

Explicação passo a passo:

$27^{x^{2} +1} =9^{5x} \\\\ (3^{3} )^{x^{2} +1} =(3^{2} )^{5x} \\\\ 3^{3x^{2} +3} =3^{10x} \\\\ 3x^{2} +3=10x \\\\ 3x^{2} -10x+3=0$

$Fo\´rmula\ de\ Bhaskara:$

$x=\frac{-(-10)\pm\sqrt{(-10)^{2}-4\times 3\times3} }{2\times3} \\\\ x=\frac{10\pm\sqrt{100-12\times3} }{6} \\\\ x=\frac{10\pm\sqrt{100-36} }{6} \\\\ x=\frac{10\pm\sqrt{64} }{6} \\\\ x=\frac{10\pm\sqrt{8^{2} } }{6} \\\\ x=\frac{10\pm8}{6} \\\\ x=\frac{5\pm4}{3} \\\\\\ x'=\frac{5-4}{3} \\\\\\ \bold{x'=\frac{1}{3}} \\\\\\\\ x''=\frac{5+4}{3} \\\\\\ x''=\frac{9}{3} \\\\\\ \bold{x'' = 3}$