Question

URGENTE! RESOLUÇÃO / Seja a equação exponencial 9^x+3= (1/81)^x
Assinale a alternativa que contém a
solução da equação exponencial dada.
a) x= 5
b) x= -5
c) x= 8
d) X= -9
e) N.R.A
RESOLUÇÃO, por favor é urgente

Answer 1

$\text{Si la pregunta es: } 9^{x+3}=(1/81)^x\\\\ (3^2)^{x+3}=(3^{-4})^x\\ \\ 3^{2x+6}=3^{-4x}\\ \\ 2x+6=-4x\\ \\ 10x=-6$

ou

$9^x+3=(1/81)^x \text{ ent\~ao}\\ \\ 3^{2x}+3=3^{-4x}\\ \\ 3^{4x}(3^{2x}+3)=1\\ \\ \text{Seja }z=3^{2x} \text{ ent\~ao:}\\ \\ z^2(z+3)=1\\ \\ z^3+3z^2-1=0\\ \\ \text{si }z=y-1\longrightarrow z^3+3z^2-1=0 \iff y^3-3y+1=0\\ \text{haciendo }y=2\cos \theta: 2(4\cos^3\theta-3\cos\theta)+1=0$

$2\cos(3\theta)+1=0\\ \cos(3\theta) = -1/2\\ 3\theta\in\{2\pi/3,4\pi/3, 8\pi/3\}\\ \\ \theta\in\{2\pi/9,4\pi/9, 8\pi/9\}\\ \\ \text{Entonces:}\\ \displaystyle y\in \left\{2\cos\frac{2\pi}{9},2\cos\frac{4\pi}{9},2\cos\frac{8\pi}{9} \right\}$

$\displaystyle z\in \left\{2\cos\frac{2\pi}{9}-1\,,\,2\cos\frac{4\pi}{9}-1\,,\,2\cos\frac{8\pi}{9}-1 \right\}\\ \\ 3^{2x}\in \left\{2\cos\frac{2\pi}{9}-1\,,\,2\cos\frac{4\pi}{9}-1\,,\,2\cos\frac{8\pi}{9}-1 \right\}$

$\displaystyle2x\in \left\{\log_3\left(2\cos\frac{2\pi}{9}-1\right)\,,\,\log_3\left(2\cos\frac{4\pi}{9}-1\right)\,,\,\log_3\left(2\cos\frac{8\pi}{9}-1\right)\right\}$

$x\in \left\{\log_3\sqrt{2\cos\frac{2\pi}{9}-1}\,,\,\log_3\sqrt{2\cos\frac{4\pi}{9}-1}\,,\,\log_3\sqrt{2\cos\frac{8\pi}{9}-1}\right\}$

Respuesta (e)