Question

Equação exponencial
(4^x)^x-1 =16

Answer 1

Olá, tudo bem?

$(4^x)^{x-1}=16\rightarrow 4^{x^{2}-x}=4^{2}\rightarrow x^{2}-x-2=0\quad \underrightarrow{Bhaskara} \\\\ x=\dfrac{-(-1)\pm \sqrt{(-1)^{2}-4\times 1\times (-2)}}{2\times 1}\rightarrow x=\dfrac{1\pm \sqrt{9}}{2}\rightarrow \\\\ x=\dfrac{1\pm 3}{2}\rightarrow \boxed{x_{1}=-1}\,\,\text{ou}\,\, \boxed{x_{2}=2}\,\,\text{(respostas finais)}$

Bons Estudos e até mais!!