algumas equações exponenciais podem ser resolvidas, com maior facilidade, atraves do uso de artificio. com o uso de artificio, a solução da seguinte equação: 4x-2x.6+8=0. ME AJUDEM
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Boa tarde Belinha!
Solução!
O artificio aqui consiste em transformar a exponencial em uma equação do segundo grau,feito isso analisar se as raízes satisfaz o conjunto verdade.
![4^{x}+2^{x}.6+8=0\\\\\\
2^{2}^{(x)}+2^{x}.6+8=0\\\\\\\
(2^{x})^{2}+2^{x}.6+8=0\\\\\\\
Artificio!\\\\\\
2^{x}=y](https://tex.z-dn.net/?f=4%5E%7Bx%7D%2B2%5E%7Bx%7D.6%2B8%3D0%5C%5C%5C%5C%5C%5C%0A2%5E%7B2%7D%5E%7B%28x%29%7D%2B2%5E%7Bx%7D.6%2B8%3D0%5C%5C%5C%5C%5C%5C%5C%0A++%282%5E%7Bx%7D%29%5E%7B2%7D%2B2%5E%7Bx%7D.6%2B8%3D0%5C%5C%5C%5C%5C%5C%5C+%0AArtificio%21%5C%5C%5C%5C%5C%5C%0A2%5E%7Bx%7D%3Dy+++ "4^{x}+2^{x}.6+8=0\\\\\\
2^{2}^{(x)}+2^{x}.6+8=0\\\\\\\
(2^{x})^{2}+2^{x}.6+8=0\\\\\\\
Artificio!\\\\\\
2^{x}=y")
![(2^{x})^{2}+2^{x}.6+8=0\\\\\\\
y^{2}+y.6+8=0\\\\\\\
y^{2}+6y+8=0\\\\\\\
Eq~~do~~segundo~~grau,vamos~~aplicar~~a~~formula~~de~~Bhaskara.\\\\\\\\\\
\boxed{y= \dfrac{-b\pm \sqrt{b^{2}-4.a.c } }{2.a}}\\\\\\
y= \dfrac{-(-6)\pm \sqrt{(-6)^{2} -4.1.8 } }{2.1}\\\\\\
y= \dfrac{6\pm \sqrt{36 -32} }{2}\\\\\\
y= \dfrac{6\pm \sqrt{4} }{2}\\\\\\
y= \dfrac{6\pm 2}{2}\\\\\\
Raizes!\\\\\\](https://tex.z-dn.net/?f=%282%5E%7Bx%7D%29%5E%7B2%7D%2B2%5E%7Bx%7D.6%2B8%3D0%5C%5C%5C%5C%5C%5C%5C%0Ay%5E%7B2%7D%2By.6%2B8%3D0%5C%5C%5C%5C%5C%5C%5C%0A+y%5E%7B2%7D%2B6y%2B8%3D0%5C%5C%5C%5C%5C%5C%5C%0A%0AEq%7E%7Edo%7E%7Esegundo%7E%7Egrau%2Cvamos%7E%7Eaplicar%7E%7Ea%7E%7Eformula%7E%7Ede%7E%7EBhaskara.%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7By%3D+%5Cdfrac%7B-b%5Cpm+%5Csqrt%7Bb%5E%7B2%7D-4.a.c+%7D+%7D%7B2.a%7D%7D%5C%5C%5C%5C%5C%5C%0A%0Ay%3D+%5Cdfrac%7B-%28-6%29%5Cpm+%5Csqrt%7B%28-6%29%5E%7B2%7D+-4.1.8+%7D+%7D%7B2.1%7D%5C%5C%5C%5C%5C%5C%0A%0Ay%3D+%5Cdfrac%7B6%5Cpm+%5Csqrt%7B36+-32%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0A%0Ay%3D+%5Cdfrac%7B6%5Cpm+%5Csqrt%7B4%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0A%0Ay%3D+%5Cdfrac%7B6%5Cpm+2%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0ARaizes%21%5C%5C%5C%5C%5C%5C%0A+%0A%0A%0A++ "(2^{x})^{2}+2^{x}.6+8=0\\\\\\\
y^{2}+y.6+8=0\\\\\\\
y^{2}+6y+8=0\\\\\\\
Eq~~do~~segundo~~grau,vamos~~aplicar~~a~~formula~~de~~Bhaskara.\\\\\\\\\\
\boxed{y= \dfrac{-b\pm \sqrt{b^{2}-4.a.c } }{2.a}}\\\\\\
y= \dfrac{-(-6)\pm \sqrt{(-6)^{2} -4.1.8 } }{2.1}\\\\\\
y= \dfrac{6\pm \sqrt{36 -32} }{2}\\\\\\
y= \dfrac{6\pm \sqrt{4} }{2}\\\\\\
y= \dfrac{6\pm 2}{2}\\\\\\
Raizes!\\\\\\")
![y_{1} = \dfrac{6+2}{2} = \dfrac{8}{2}=4\\\\\\\
y_{2} = \dfrac{6-2}{2} = \dfrac{4}{2}=2](https://tex.z-dn.net/?f=y_%7B1%7D+%3D+%5Cdfrac%7B6%2B2%7D%7B2%7D+%3D+%5Cdfrac%7B8%7D%7B2%7D%3D4%5C%5C%5C%5C%5C%5C%5C%0A+y_%7B2%7D+%3D+%5Cdfrac%7B6-2%7D%7B2%7D+%3D+%5Cdfrac%7B4%7D%7B2%7D%3D2 "y_{1} = \dfrac{6+2}{2} = \dfrac{8}{2}=4\\\\\\\
y_{2} = \dfrac{6-2}{2} = \dfrac{4}{2}=2")
![Retomando~~ o~~ modelo ~~que ~~usamos~~ como ~~artificio,vamos\\\\\ determinar~~ o~~ valor ~~de~~x.\\\\\\\
2^{x}=y_{1} \\\\\\
2^{x}=4\\\\\\
2^{x}=2^{2}\\\\\\
\boxed{x=2}\\\\\\\\\\
2^{x}=y_{2} \\\\\\
2^{x}=2^{1} \\\\\\
\boxed{x=1}\\\\\\\\\\
\boxed{Resposta:S=\left \{ 1,2 \right \}}](https://tex.z-dn.net/?f=Retomando%7E%7E+o%7E%7E+modelo+%7E%7Eque+%7E%7Eusamos%7E%7E+como+%7E%7Eartificio%2Cvamos%5C%5C%5C%5C%5C+determinar%7E%7E+o%7E%7E+valor+%7E%7Ede%7E%7Ex.%5C%5C%5C%5C%5C%5C%5C%0A2%5E%7Bx%7D%3Dy_%7B1%7D+%5C%5C%5C%5C%5C%5C%0A2%5E%7Bx%7D%3D4%5C%5C%5C%5C%5C%5C%0A2%5E%7Bx%7D%3D2%5E%7B2%7D%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Bx%3D2%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A+2%5E%7Bx%7D%3Dy_%7B2%7D+%5C%5C%5C%5C%5C%5C%0A2%5E%7Bx%7D%3D2%5E%7B1%7D++%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Bx%3D1%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BResposta%3AS%3D%5Cleft+%5C%7B+1%2C2+%5Cright+%5C%7D%7D "Retomando~~ o~~ modelo ~~que ~~usamos~~ como ~~artificio,vamos\\\\\ determinar~~ o~~ valor ~~de~~x.\\\\\\\
2^{x}=y_{1} \\\\\\
2^{x}=4\\\\\\
2^{x}=2^{2}\\\\\\
\boxed{x=2}\\\\\\\\\\
2^{x}=y_{2} \\\\\\
2^{x}=2^{1} \\\\\\
\boxed{x=1}\\\\\\\\\\
\boxed{Resposta:S=\left \{ 1,2 \right \}}")
Boa tarde!
Bons estudos!
Solução!
O artificio aqui consiste em transformar a exponencial em uma equação do segundo grau,feito isso analisar se as raízes satisfaz o conjunto verdade.
Boa tarde!
Bons estudos!
belinhahh1:
porem é 4x-2x.6=8=0
Esta falando que é uma exponencial!
sim. e isso interfere no sinal ?
Qual sinal?
De menos em frente ao 2
Não!
Então continua menos?
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