Question

Se x é um número real, resolva a equação exponencial 3^2x + 3^x + 1 = 18

Answer 1

$\begin{array}{l}\qquad~~3^{2x}+3^x+1=18~~\Longleftrightarrow\\\\\Longleftrightarrow~~(3^x)^2+3^x+1-18=0~~\Longleftrightarrow\\\\\qquad\quad~~ Artificio: 3^x=a\\\\\Longleftrightarrow~~a^2+a-17=0~~\Longleftrightarrow\\\\\Longleftrightarrow~~a=\dfrac{-\,1\pm\sqrt{1^2-4\cdot1\cdot(-\,17)}}{2\cdot1}~~\Longleftrightarrow\\\\\Longleftrightarrow~~ a=\dfrac{-\,1\pm\sqrt{1+68}}{2}~~\Longleftrightarrow\\\\\Longleftrightarrow~~a=\dfrac{-\,1\pm\sqrt{69}}{2}.\end{array}$

Voltando à variável antiga:

$\begin{array}{l}3^x=\dfrac{-\,1+\sqrt{69}}{2}~~ou~~3^x=\dfrac{-\,1-\sqrt{69}}{2}.\end{array}$

Definição (logaritmo): $a^c=b\Longleftrightarrow c= log_a\,b,~1\neq a > 0~e~b > 0$. Portanto:

$\begin{array}{l}x=log_3\,\bigg(\dfrac{-\,1+\sqrt{69}}{2}\bigg)~~ou~~x=log_3\,\bigg(\dfrac{-\,1-\sqrt{69}}{2}\bigg).\end{array}$

Dado que $b > 0$ (logaritmando é positivo), descartaremos a segunda raiz pelo fato de que $\frac{-\,1-\sqrt{69}}{2} < 0$. Assim,

$\boxed{\begin{array}{l}x=log_3\,\bigg(\dfrac{-\,1+\sqrt{69}}{2}\bigg).\end{array}}$