Question

Determine o conjunto solução da seguinte equação

$\Large\boxed{ \dfrac{3^{x+5}}{18+3^{x+3}}=3^{x+1}+2}$

Answer 1

$\frac{3^{x+5}}{18+3^{x+3}}=3^{x+1}+2\\\\\frac{3^x.3^5}{18+3^x.3^3}}=3^x.3+2\\\\\\3^x=y\\\\\\\frac{243y}{18+27y}=3y+2\\\\243y=(3y+2).(18+27y)\\243y=54y+81y^2+36+54y\\243y=81y^2+108y+36\\81y^2-135y+36=0$

Δ = (-135)² - 4.81.36
Δ = 18.225 - 11.664
Δ = 6.561

$x=\frac{-(-135)\pm\sqrt{6.561}}{2.81}\\\\x=\frac{135+81}{162}\\\\x=\frac{216}{162}\\\\\boxed{x=\frac{4}{3}}\\\\\\x'=\frac{135-81}{162}\\\\x'=\frac{54}{162}\\\\\\\boxed{x'=\frac{1}{3}}$


$3^x=\frac{4}{3}\\\\log(3^x)=log(\frac{4}{3})\\\\x.log(3)=log(4)-log(3)\\0,477x=0,602-0,477\\0,477x=0,125\\\\x=\frac{0,125}{0,477}\\\\\boxed{x\approx 0,262}$


$3^x=\frac{1}{3}\\3^x=3^{-1}\\\\\boxed{x=-1}$


x vale aproximadamente 0,262 ou x é igual a -1

Answer 2

Resposta:

3^{x+5}/18+3^{x+3} = 3^{x+1} + 2

3^{x}•3^{5}/18+3^{x}•3^{3} = 3^{x} • 3 + 2

3^{x}•243/18+3^{x}•27 = 3^{x} • 3 + 2

t = 3^{x}

t•243/18+t•27 = t • 3 + 2

t•243/18+27t = 3t + 2

t•243/9(2+3t) = 3t + 2

t•27/2+3t = 3t + 2

27t/2+3t = 3t + 2

27t = (3t + 2) • (2 + 3t)

27t - (3t + 2) • (2 + 3t) = 0

27t - (3t + 2)^{2} = 0

27t - (9t^{2} + 12t + 4) = 0

27t - 9t^{2} - 12t - 4 = 0

15t - 9t^{2} - 4 = 0

- 9t^{2} + 15t - 4 = 0

9t^{2} - 15t + 4 = 0

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↔Função Quadrática↔

ax² + bx + c = 0 ➡ x = -b±√{b² - 4ac}/2a

t = -(- 15)±√{(- 15)² - 4 • 9 • 4}/2•9

t = 15±√{225 - 144}/18

t = 15±√81/18

t = 15±9/18

t = 15+9/18 = 4/3

t = 15-9/18 = 1/3

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t = 4/3

t = 1/3

⬆t = 3^{x}

3^{x} = 4/3 ➡ = 2log₃(2) - 1 ➡ ≈ 0,26186

3^{x} = 1/3 ➡ = - 1

S {x₁ = - 1 , x₂ ≈ 0,26186}