Question

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Answer 1

$\sqrt[x+4]{ 2^{3x} + 8 } = 2^{x-5}$

$2^{ \frac{3x}{x+4} + 2^{ \frac{3}{x+4} } } = 2^{x-5}$


$\frac{3x}{x+4} Log(2) + \frac{3}{x+4} Log(2) = x-5Log(2)$


$\frac{3x}{x+4} + \frac{3}{x+4} = x-5$

$3x + \frac{3.(x+4)}{x+4} = x-5.(x+4)$


$3x + \frac{3x+12}{x+4} = x^2+4x-5x-20$

$3x .(x+4) + 3x+12 = x^2-x-20 .(x+4)$

$3x^2 + 12x + 3x+12 = x^3 -x^2 -20x+4x^2-4x-80$

$3x^2 + 12x + 3x+12 = x^3 -x^2 -20x+4x^2-4x-80$

-x³ + 39x + 92 = 0

Raízes desse polinômio:

$x = (-4) x' = (2+3 \sqrt{3}) ou (2-3 \sqrt{3})$

Answer 2

$\sqrt[x+4]{ 2^{3x} + 2^{3} } = 2^{x-5}$

$2^{ \frac{3x}{x+4} }+ 2^{ \frac{3}{x+4} } = 2^{x-5}$

$log 2^{ \frac{3x}{x+4} } + log 2^{ \frac{3}{x+4} } = log 2^{x-5}$

$log 2^{ \frac{3x}{x+4} }. 2^{ \frac{3}{x+4} } = log 2^{x-5}$

$\frac{3x+3}{x+4} log2 = x-5 (log2)$

3x+3 = x² -5x+ 4x -20 
x² -4x -23 = 0 , a = 1 b = -4  e c = -23

Δ= (-4)² - 4.1.-23
Δ = 108

x' = $\frac{-b+ \sqrt{delta} }{2a} = \frac{4+6 \sqrt{3} }{2} = 2+3 \sqrt{3}$

x'' = 2-3√3