Determine o conjunto solução da seguinte equação exponencial:
![2^x^-^3+2^x^-^1+2^x=52 2^x^-^3+2^x^-^1+2^x=52](https://tex.z-dn.net/?f=2%5Ex%5E-%5E3%2B2%5Ex%5E-%5E1%2B2%5Ex%3D52)
O produto das soluções da equação
é:
a) 0
b) 1
c) 4
d) 5
e) 6
Soluções para a tarefa
Respondido por
7
1) ![2^{x-3}+2^{x-1}+2^{x}=52 2^{x-3}+2^{x-1}+2^{x}=52](https://tex.z-dn.net/?f=2%5E%7Bx-3%7D%2B2%5E%7Bx-1%7D%2B2%5E%7Bx%7D%3D52)
![2^{x}\cdot2^{-3}+2^{x}=\cdot2^{-1}+2^{x}=52 2^{x}\cdot2^{-3}+2^{x}=\cdot2^{-1}+2^{x}=52](https://tex.z-dn.net/?f=2%5E%7Bx%7D%5Ccdot2%5E%7B-3%7D%2B2%5E%7Bx%7D%3D%5Ccdot2%5E%7B-1%7D%2B2%5E%7Bx%7D%3D52)
![2^{x}\cdot(2^{-3}+2^{-1}+1)=52 2^{x}\cdot(2^{-3}+2^{-1}+1)=52](https://tex.z-dn.net/?f=2%5E%7Bx%7D%5Ccdot%282%5E%7B-3%7D%2B2%5E%7B-1%7D%2B1%29%3D52)
![2^{x}\cdot(\frac{1}{8}+\frac{1}{2}+1)=52 2^{x}\cdot(\frac{1}{8}+\frac{1}{2}+1)=52](https://tex.z-dn.net/?f=2%5E%7Bx%7D%5Ccdot%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B1%7D%7B2%7D%2B1%29%3D52)
![2^{x}\cdot1,625=52 2^{x}\cdot1,625=52](https://tex.z-dn.net/?f=2%5E%7Bx%7D%5Ccdot1%2C625%3D52)
![2^{x}=\dfrac{52}{1,625} 2^{x}=\dfrac{52}{1,625}](https://tex.z-dn.net/?f=2%5E%7Bx%7D%3D%5Cdfrac%7B52%7D%7B1%2C625%7D)
![2^{x}=32 2^{x}=32](https://tex.z-dn.net/?f=2%5E%7Bx%7D%3D32)
![2^{x}=2^5 2^{x}=2^5](https://tex.z-dn.net/?f=2%5E%7Bx%7D%3D2%5E5)
.
2)![(4^{3-x})^{2-x}=1 (4^{3-x})^{2-x}=1](https://tex.z-dn.net/?f=%284%5E%7B3-x%7D%29%5E%7B2-x%7D%3D1)
![4^{(3-x)(2-x)}=4^0 4^{(3-x)(2-x)}=4^0](https://tex.z-dn.net/?f=4%5E%7B%283-x%29%282-x%29%7D%3D4%5E0)
![4^{x^2-5x+6}=4^{0} 4^{x^2-5x+6}=4^{0}](https://tex.z-dn.net/?f=4%5E%7Bx%5E2-5x%2B6%7D%3D4%5E%7B0%7D)
![x^2-5x+6=0 x^2-5x+6=0](https://tex.z-dn.net/?f=x%5E2-5x%2B6%3D0)
![P=\dfrac{c}{a} P=\dfrac{c}{a}](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7Bc%7D%7Ba%7D)
![P=\dfrac{6}{1} P=\dfrac{6}{1}](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7B6%7D%7B1%7D)
.
Letra E
2)
Letra E
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