Question

Resolva em $\mathbb{R}$ a seguinte equação:
$e^x+e^{2x}=e^{3x}$

Answer 1

$\large\sf{} {e}^{x} + {e}^{2x} = {e}^{3x} \\ \\ \large\sf{} {e}^{x} + {e}^{2x} - {e}^{3x} = 0 \\ \large\sf{} {e}^{x} \times\left(1 + {e}^{x} - {x}^{2x} \right) = 0 \\ \large\sf{} {e}^{x} = 0 \\ \large\sf{}1 + {e}^{x} - {e}^{2x} = 0 \\ \\ \large\sf{} \emptyset \\ \large\sf{}x = in\left( \dfrac{1 + \sqrt{5} }{2} \right) \\ \\ \large\sf{}x = in\left( \dfrac{1 + \sqrt{5} }{2} \right) \\ \\ {\large\boxed{\boxed{  { \large \sf x = in\left( \dfrac{1 + \sqrt{5} }{2} \right)}}}}$

${\large\boxed{\boxed{  { \large \sf Bons~Estudos }}}}$

Answer 2

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\sf e^x + e^{2x} = e^{3x}$

$\sf e^x + e^{2x} - e^{3x} = 0$

$\sf e^x\:.\:(1 + e^{x} - e^{2x}) = 0$

$\boxed{\sf e^x = 0 \Leftrightarrow S = \{\}}$

$\sf e^{2x} - e^x - 1 = 0$

$\sf e^x = y$

$\sf y^2 - y - 1 = 0$

$\sf a = 1 \Leftrightarrow b = -1 \Leftrightarrow c = -1$

$\sf \Delta = b^2 - 4.a.c$

$\sf \Delta = (-1)^2 - 4.1.(-1)$

$\sf \Delta = 1 + 4$

$\sf \Delta = 5$

$\sf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{1 \pm \sqrt{5}}{2} \rightarrow \begin{cases}\sf{y' = \dfrac{1 + \sqrt{5}}{2}}\\\\\sf{y'' = \dfrac{1 - \sqrt{5}}{2}}\end{cases}}$

$\boxed{\sf e^x = \dfrac{1 - \sqrt{5}}{2} \Leftrightarrow S = \{\}}$

$\sf e^x = \dfrac{1 + \sqrt{5}}{2}$

$\sf ln\:e^x = ln\left(\dfrac{1 + \sqrt{5}}{2}\right)$

$\sf S = \left\{ln\left(\dfrac{1 + \sqrt{5}}{2}\right)\right\}$