Sendo a = [(2^48) + (4^22) - (2^46)]/ 2 * (8^15) , obtenha o valor de (4a)^-1 (Sem o uso de calculadoras)
Respostas
respondido por:
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Podemos perceber que 4 = 2.2 = 2²
e 8 = 4.2 = 2.2.2 = 2³
Substituindo na equação, temos:
a=![\frac{(2^{48}+(2^{2})^{22}-2^{46})}{2.(2^{3})^{15}} \frac{(2^{48}+(2^{2})^{22}-2^{46})}{2.(2^{3})^{15}}](https://tex.z-dn.net/?f=+%5Cfrac%7B%282%5E%7B48%7D%2B%282%5E%7B2%7D%29%5E%7B22%7D-2%5E%7B46%7D%29%7D%7B2.%282%5E%7B3%7D%29%5E%7B15%7D%7D+)
Pelas propriedades da potenciação:
![2^{x}.2^{y}=2^{x+y} \\ \frac{2^{x}}{2^{y}}=2^{x-y} 2^{x}.2^{y}=2^{x+y} \\ \frac{2^{x}}{2^{y}}=2^{x-y}](https://tex.z-dn.net/?f=2%5E%7Bx%7D.2%5E%7By%7D%3D2%5E%7Bx%2By%7D+%5C%5C+%5Cfrac%7B2%5E%7Bx%7D%7D%7B2%5E%7By%7D%7D%3D2%5E%7Bx-y%7D+)
![(2^{a})^{b}=2^{a.b} (2^{a})^{b}=2^{a.b}](https://tex.z-dn.net/?f=%282%5E%7Ba%7D%29%5E%7Bb%7D%3D2%5E%7Ba.b%7D)
Então a equação fica na forma:
![\frac{(2^{48}+(2^{2.22})-2^{46})}{2.(2^{3.15})}=\frac{(2^{48}+2^{44}-2^{46})}{2.(2^{45})}=\frac{(2^{48}+2^{44}-2^{46})}{2^{46}} \frac{(2^{48}+(2^{2.22})-2^{46})}{2.(2^{3.15})}=\frac{(2^{48}+2^{44}-2^{46})}{2.(2^{45})}=\frac{(2^{48}+2^{44}-2^{46})}{2^{46}}](https://tex.z-dn.net/?f=%5Cfrac%7B%282%5E%7B48%7D%2B%282%5E%7B2.22%7D%29-2%5E%7B46%7D%29%7D%7B2.%282%5E%7B3.15%7D%29%7D%3D%5Cfrac%7B%282%5E%7B48%7D%2B2%5E%7B44%7D-2%5E%7B46%7D%29%7D%7B2.%282%5E%7B45%7D%29%7D%3D%5Cfrac%7B%282%5E%7B48%7D%2B2%5E%7B44%7D-2%5E%7B46%7D%29%7D%7B2%5E%7B46%7D%7D)
Colocando 2^46 em evidência:
![a=\frac{2^{46}(2^{2}+2^{-2}-2)}{2^{46}}=2^{2}+ \frac{1}{2^{2}}-2=4+ \frac{1}{4}-2=2+ \frac{1}{4} \\ \\ = \frac{8}{4}+ \frac{1}{4}= \frac{9}{4} a=\frac{2^{46}(2^{2}+2^{-2}-2)}{2^{46}}=2^{2}+ \frac{1}{2^{2}}-2=4+ \frac{1}{4}-2=2+ \frac{1}{4} \\ \\ = \frac{8}{4}+ \frac{1}{4}= \frac{9}{4}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B2%5E%7B46%7D%282%5E%7B2%7D%2B2%5E%7B-2%7D-2%29%7D%7B2%5E%7B46%7D%7D%3D2%5E%7B2%7D%2B+%5Cfrac%7B1%7D%7B2%5E%7B2%7D%7D-2%3D4%2B+%5Cfrac%7B1%7D%7B4%7D-2%3D2%2B+%5Cfrac%7B1%7D%7B4%7D+%5C%5C++%5C%5C+%3D+%5Cfrac%7B8%7D%7B4%7D%2B+%5Cfrac%7B1%7D%7B4%7D%3D+%5Cfrac%7B9%7D%7B4%7D+)
Portanto, o valor de (4a)^-1 é:
![(4a)^{-1}= \frac{1}{4a} = \frac{1}{ 4.\frac{9}{4} } = \frac{1}{9} (4a)^{-1}= \frac{1}{4a} = \frac{1}{ 4.\frac{9}{4} } = \frac{1}{9}](https://tex.z-dn.net/?f=%284a%29%5E%7B-1%7D%3D+%5Cfrac%7B1%7D%7B4a%7D+%3D+%5Cfrac%7B1%7D%7B+4.%5Cfrac%7B9%7D%7B4%7D+%7D+%3D+%5Cfrac%7B1%7D%7B9%7D+)
e 8 = 4.2 = 2.2.2 = 2³
Substituindo na equação, temos:
a=
Pelas propriedades da potenciação:
Então a equação fica na forma:
Colocando 2^46 em evidência:
Portanto, o valor de (4a)^-1 é:
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