• Matéria: Matemática
  • Autor: jefter15alves
  • Perguntado 3 anos atrás


 {5}^{0.444x}  + {0.3}^{ \sqrt{2} }  =  {0.2}^{ \sqrt[5]{3} }  + 1.2
precisamos descobrir o valor de x ​

Respostas

respondido por: VitiableIndonesia
0

 {5}^{0.44 4x}  + 0.3 {}^{ \sqrt{2} }  = 0.2 {}^{ \sqrt[5]{3} }  + 1.2 \\ 5 {}^{0.444x}  + \left({  \frac{3}{10}  }\right) {}^{ \sqrt{2} }  = 0.2 {}^{ \sqrt[5]{3} }  + 1.2 \\  {5}^{0.444x}  + \left({  \frac{3}{10}  }\right) {}^{ \sqrt{2} }  = \left({  \frac{2}{10}  }\right) {}^{ \sqrt[5]{3} }  +  \frac{12}{10}

Simplifique as duas últimas frações

 {5}^{0.444x}  + \left({  \frac{3}{10}  }\right) {}^{ \sqrt{2} }  =  \left({ \frac{1}{5} }\right) {}^{ \sqrt[5]{3} }  +  \frac{6}{5}  \\  \\  {5}^{0.444x}  +  \frac{ {3}^{ \sqrt{2} } }{ {10}^{ \sqrt{2} } }  =  \frac{1}{ {5}^{ \sqrt[5]{3} } }  +  \frac{6}{5}  \\  \\  \frac{ {5}^{0.444x} }{1}  +  \frac{ {3}^{ \sqrt{2} } }{ {10}^{ \sqrt{2} } }  =  \frac{1}{ {5}^{ \sqrt[5]{3} } }  +  \frac{6}{5}  \\  \\  \frac{ {10}^{ \sqrt{2} }  \times  {5}^{0.444x} }{ {10}^{ \sqrt[]{2} } }  +  \frac{ {3}^{ \sqrt{2} } }{ {10}^{ \sqrt{2} } }  =  \frac{1}{ {5}^{ \sqrt[5]{3} } }  +  \frac{6}{5}

Multiplique a equação por 5 \times  {10}^{ \sqrt{2} }  \times  {5}^{ \sqrt[5]{3} }

5 \times  {10}^{ \sqrt{2} } \times  {5}^{ \sqrt[5]{3} }   \times  \frac{ {10}^{ \sqrt{2} }  \times  {5}^{0.444x} }{ {10}^{ \sqrt[]{2} } }  +  \frac{ {3}^{ \sqrt{2} } }{ {10}^{ \sqrt{2} } }  = 5 \times  {10}^{ \sqrt{2} }  \times  {5}^{ \sqrt[5]{3} } \times  \left({  \frac{1}{ {5}^{ \sqrt[5]{3} } }  +  \frac{6}{5}  }\right) \\  \\ 5 \times  {\cancel{ {10}^{ \sqrt{2} } } } \times  {5}^{ \sqrt[5]{3} }   \times  \frac{ {10}^{ \sqrt{2} }  \times  {5}^{0.444x} }{ {\cancel{ {10}^{ \sqrt{2} } } } }  +  \frac{ {3}^{ \sqrt{2} } }{ {10}^{ \sqrt{2} } }  = 5 \times  {10}^{ \sqrt{2} }  \times  {5}^{ \sqrt[5]{3} } \times  \left({  \frac{1}{ {5}^{ \sqrt[5]{3} } }  +  \frac{6}{5}  }\right) \\  \\ 5 \times {5}^{ \sqrt[5]{3} }  \times \left({  {10}^{ \sqrt{2} }  \times  {5}^{0.444x}   +  {3}^{ \sqrt{2} } }\right) = 5 \times  {10}^{ \sqrt{2} }  \times  {5}^{ \sqrt[5]{3}  }  \times  \frac{1}{ {5}^{ \sqrt[5]{3} } }  + \cancel{5} \times  {10}^{ \sqrt{2} }  \times 5 {}^{ \sqrt[5]{3} }  \times  \frac{6}{ \cancel{5}}  \\  \\  {5}^{1 +  \sqrt[5]{3} }  \times \left({  {10}^{ \sqrt{2} }   \times  {5}^{0.444x}  +  {3}^{ \sqrt{2} } }\right) = 5 \times  {10}^{ \sqrt{2} }  +  {10}^{ \sqrt{2} }  \times  {5}^{ \sqrt[5]{3} }  \times 6

 {10}^{ \sqrt{2} }  \times  {5}^{1 +  \sqrt[5]{3} + 0.444x }  +  {5}^{1 +  \sqrt[5]{3}  }  \times  {3}^{ \sqrt{2} }  = 5 \times  {10}^{ \sqrt{2} }  + 6 \times  {10}^{ \sqrt{2} }  \times  {5}^{ \sqrt[5]{3} }  \\  \\  {5}^{1 +  \sqrt[5]{3} + 0.444x }  =  \frac{5 \times  {10}^{ \sqrt{2} } + 6 \times  {10}^{ \sqrt{2} } \times  {5}^{ \sqrt[5]{3} }  -  {5}^{1 +  \sqrt[5]{3} } \times  {3}^{ \sqrt{2} }    }{ {10}^{ \sqrt{2} } }

 log_{5}\left({  {5}^{1 +  \sqrt[5]{3} + 0.444x }   }\right)  =  log_{5}\left({  \frac{5 \times  {10}^{ \sqrt{2}  }  + 6 \times  {10}^{ \sqrt{2} }  \times  {5}^{ \sqrt[5]{3} }  -  {5}^{1 +  \sqrt[5]{3} } \times  {3}^{ \sqrt{2} }  }{ {10}^{ \sqrt{2} } }  }\right) \\  \\ 1 +  \sqrt[5]{3}  + 0.444x =  log_{5}\left({  \frac{5 \times  {10}^{ \sqrt{2} } + 6 \times  {10}^{ \sqrt{2} } \times  {5}^{ \sqrt[5]{3} }  -  {5}^{1 +  \sqrt[5]{3} } \times  {3}^{ \sqrt{2} }    }{ {10}^{ \sqrt{2} } }  }\right) \\  \\

Faça a divisão dos membros da equação por 0,444

x =  log_{5}\left({  \frac{5 \times  {10}^{ \sqrt{2} }6 \times  {10}^{ \sqrt{2} } \times  {5}^{ \sqrt[5]{3} }   -  {5}^{1 +  \sqrt[5]{3} }  \times  {3}^{ \sqrt{2} }  }{ {10}^{ \sqrt{2} } }  }\right)  \times  \frac{250}{111}  -  \frac{250}{111}  -  \frac{250 \sqrt[5]{3} }{111}  \\  \\\color{green}\begin{gathered} \boxed{\begin{array}{lr}\large\sf\:x≈0,198591 \large \sf \large \sf  \: \end{array}}\end{gathered}

\mathcal{Bons \: estudos } \\ \displaystyle\int_ \empty ^ \mathbb{C}     \frac{ - b \: ± \:  \sqrt{ {b}^{2} - 4 \times a \times c } }{2 \times a} d{ t } \boxed{ \boxed{ \mathbb{\displaystyle\Re}\sf{ \gamma  \alpha }\tt{ \pi}\bf{ \nabla}}}

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