Question

Como resolver equação exponencial?

$17^{x-4x} =1$

e

$5^{ x^{2} +2x} =1$

Answer 1

$17^{x-4x}=1$

$17^{-3x}=17^{0}$

$-3x=0$

$x=0$

Se for $17^{x^2-4x}=1~~\Rightarrow~~17^{x^2-4x}=17^{0}$.

$x^2-4x=0~~\Rightarrow~~x(x-4)=0~~\Rightarrow~~x=4$ ou $x=0$.



$5^{x^2+2x}=1$

$5^{x^2+2x}=5^{0}$

$x^2+2x=0~~\Rightarrow~~x(x+2)=0~~\Rightarrow~~x=0$ ou $x=-2$.