Question

Equação exponencial:

Ache o valor de x:

$4^{x+2} + 16^{x} + 8$

Answer 1

Oie, tudo bom?

$4 {}^{x + 2} + 16 {}^{x} = 8 \\ 4 {}^{x} \: . \: 4 {}^{2} + (4 {}^{2} ) {}^{x} = 8 \\ 4 {}^{x} \: . \: 16 + (4 {}^{x}) {}^{2} = 8 \\ \boxed{ 4 {}^{x} = t } \\ t \: . \: 16 + t {}^{2} = 8 \\ 16t + t {}^{2} = 8 \\ 16t + t {}^{2} - 8 = 0 \\ t {}^{2} + 16t - 8 = 0 \\ \boxed{a = 1 \: , \: b = 16 \: , \: c = - 8} \\ x = \frac{ - b± \sqrt{b {}^{2} - 4ac} }{2a} \\ t = \frac{ - 16± \sqrt{16 {}^{2} - 4 \: . \: 1 \: . \: ( - 8)} }{2 \: . \: 1} \\ t = \frac{ - 16± \sqrt{256 + 32} }{2} \\ t = \frac{ - 16± \sqrt{288} }{2} \\ t= \frac{ - 16±12 \sqrt{2} }{2} \\ t= \frac{ - 16 + 12 \sqrt{2} }{2} = \boxed{ - 8 + 6 \sqrt{2} } \\ t = \frac{ - 16 - 12 \sqrt{2} }{2} = \boxed{ - 8 - 6 \sqrt{2} } \\ \boxed{t = 4 {}^{x} } \\ 4 {}^{x} = - 8 + 6 \sqrt{2} \\ log_{4}(4 {}^{x} ) = log_{4}( - 8 + 6 \sqrt{2} ) \\ x = log_{2 {}^{2} }( - 8 + 6 \sqrt{2} ) \\ \boxed{x = \frac{1}{2} \: . \: log_{2}( - 8 + 6 \sqrt{2} ) } \\ 4 {}^{x} = - 8 - 6 \sqrt{2} \\ \boxed{x∉\mathbb{R}}$

$\boxed{S = \left \{\frac{1}{2} \: . \: log_{2}( - 8 + 6 \sqrt{2} )\right \}}$

Att. NLE Top Shotta

Answer 2

Resposta:

$4^{x\cdot \:2}+16^x=8\\\\x\cdot \:2=1\\\\x=\frac{1}{2}$