PRECISO DE AJUDA, URGENTE NESTA
Anexos:
Soluções para a tarefa
Respondido por
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Boa tarde Nicolas!
Solução!
Para resolver essa equação exponencial, tem que usar um artificio para transforma-la em uma equação do segundo grau.
![2^{2x}-9.2^{x}+8=0\\\\\
(2^{x})^{2} -9.2^{x}+8=0\\\\\
\boxed{2^{x}=y}](https://tex.z-dn.net/?f=2%5E%7B2x%7D-9.2%5E%7Bx%7D%2B8%3D0%5C%5C%5C%5C%5C%0A%282%5E%7Bx%7D%29%5E%7B2%7D+-9.2%5E%7Bx%7D%2B8%3D0%5C%5C%5C%5C%5C%0A%5Cboxed%7B2%5E%7Bx%7D%3Dy%7D+ "2^{2x}-9.2^{x}+8=0\\\\\
(2^{x})^{2} -9.2^{x}+8=0\\\\\
\boxed{2^{x}=y}")
![y^{2}-9y+8=0\\\\\\\\
y= \dfrac{9\pm \sqrt{(-9)^{2}-4.1.8} }{2.1}\\\\\\\\\
y= \dfrac{9\pm \sqrt{81-32} }{2.1}\\\\\\\\\
y= \dfrac{9\pm \sqrt{49} }{2}\\\\\\\\\
y= \dfrac{9\pm7 }{2}\\\\\\\\\
y_{1}= \dfrac{9+7}{2}= \dfrac{16}{2}=8\\\\\\
y_{2}= \dfrac{9-7}{2}= \dfrac{2}{2}=1\\\\\\](https://tex.z-dn.net/?f=y%5E%7B2%7D-9y%2B8%3D0%5C%5C%5C%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B9%5Cpm+%5Csqrt%7B%28-9%29%5E%7B2%7D-4.1.8%7D+%7D%7B2.1%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B9%5Cpm+%5Csqrt%7B81-32%7D+%7D%7B2.1%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B9%5Cpm+%5Csqrt%7B49%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%0Ay%3D+%5Cdfrac%7B9%5Cpm7+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%0Ay_%7B1%7D%3D+%5Cdfrac%7B9%2B7%7D%7B2%7D%3D+%5Cdfrac%7B16%7D%7B2%7D%3D8%5C%5C%5C%5C%5C%5C%0Ay_%7B2%7D%3D+%5Cdfrac%7B9-7%7D%7B2%7D%3D+%5Cdfrac%7B2%7D%7B2%7D%3D1%5C%5C%5C%5C%5C%5C%0A%0A%0A+++%0A "y^{2}-9y+8=0\\\\\\\\
y= \dfrac{9\pm \sqrt{(-9)^{2}-4.1.8} }{2.1}\\\\\\\\\
y= \dfrac{9\pm \sqrt{81-32} }{2.1}\\\\\\\\\
y= \dfrac{9\pm \sqrt{49} }{2}\\\\\\\\\
y= \dfrac{9\pm7 }{2}\\\\\\\\\
y_{1}= \dfrac{9+7}{2}= \dfrac{16}{2}=8\\\\\\
y_{2}= \dfrac{9-7}{2}= \dfrac{2}{2}=1\\\\\\")
![2^{x}=y_{1}\\\\\
2^{x}=8\\\\\
2^{x}=2^{3}\\\\\
\boxed{x=3}~~Soluc\~ao\\\\\\\\\\
2^{x}=y_{2}\\\\\
2^{x}=1^{0} \\\\\
\boxed{x=1} \\\\\\\\\\\
\boxed{Resposta:S=\{0,3\}}](https://tex.z-dn.net/?f=2%5E%7Bx%7D%3Dy_%7B1%7D%5C%5C%5C%5C%5C+%0A2%5E%7Bx%7D%3D8%5C%5C%5C%5C%5C%0A+2%5E%7Bx%7D%3D2%5E%7B3%7D%5C%5C%5C%5C%5C%0A%5Cboxed%7Bx%3D3%7D%7E%7ESoluc%5C%7Eao%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A2%5E%7Bx%7D%3Dy_%7B2%7D%5C%5C%5C%5C%5C+%0A2%5E%7Bx%7D%3D1%5E%7B0%7D+%5C%5C%5C%5C%5C%0A%5Cboxed%7Bx%3D1%7D+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BResposta%3AS%3D%5C%7B0%2C3%5C%7D%7D%0A%0A%0A%0A%0A "2^{x}=y_{1}\\\\\
2^{x}=8\\\\\
2^{x}=2^{3}\\\\\
\boxed{x=3}~~Soluc\~ao\\\\\\\\\\
2^{x}=y_{2}\\\\\
2^{x}=1^{0} \\\\\
\boxed{x=1} \\\\\\\\\\\
\boxed{Resposta:S=\{0,3\}}")
Boa tarde!
Bons estudos!
Solução!
Para resolver essa equação exponencial, tem que usar um artificio para transforma-la em uma equação do segundo grau.
Boa tarde!
Bons estudos!
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