Question

2^2x +3.2^x+1=8 equação exponencial ​

Answer 1

$\displaystyle \sf 2^{2x}+3\cdot 2^{x+1}=8\\\\ (2^{x})^2+3\cdot 2^x\cdot 2- 8=0 \\\\ Fa{\c c}amos: \\\\ 2^x = a \\\\ Da{\'i}}: \\\\ a^2 +6\cdot a- 8 =0 \\\\ a = \frac{-6\pm\sqrt{6^2-4\cdot1\cdot (-8)}}{2\cdot 1 }\\\\\\ a=\frac{-6\pm\sqrt{36+32}}{2} \to a = \frac{-6\pm\sqrt{4\cdot 17}}{2} \\\\\\\ a = \frac{-6\pm2\sqrt{17}}{2}\\\\\\ a = -3\pm\sqrt{17}$

como  $\sf 2^x$ é sempre maior que 0 para todo x real, então :

$\displaystyle \sf a=-3+\sqrt{17}$

Desfazendo a troca de variável :

$\displaystyle \sf 2^x=\sqrt{17}-3 \\\\\\ Aplicando\ \log_2 \ dos\ dois\ lados\ da\ igualdade:\\\\ log_2 2^x=\log_2\left[\sqrt{17}-3\right] \\\\\\ x \cdot log_22= \log_2\left[\sqrt{17}-3\right] \\\\\\ \boxed{\sf x = \log_2\left[\sqrt{17}-3\right]\ } \checkmark$

a