tatyberilo
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# sabendo que log2= 0,301, log3=0,477 e log5= 0,699 resolva as equções exponenciais: 3^x-1=5 30^x=100 20^x=3^x-2 (^) significa elevado

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vcsarah2012van

•    $3^{x-1}=5$ Aplicando logaritmo aos dois lados da equação, e usando as propriedades operatórias $\mathrm{\ell og}(3^{x-1})=\mathrm{\ell og}\,5$ ( o log da potência é o expoente multiplicado pelo log da base ) $(x-1)\cdot\mathrm{\ell og\,}3=\mathrm{\ell og}\,5\\\\ x-1=\dfrac{\mathrm{\ell og}\,5}{\mathrm{\ell og}\,3}\\\\\\ x=\dfrac{\mathrm{\ell og}\,5}{\mathrm{\ell og}\,3}+1$ Substituindo os valores dos logaritmos pelas aproximações fornecidas, $x\approx \dfrac{0,\!699}{0,\!477}+1\\\\\\ x\approx 1,\!465+1\\\\ \boxed{\begin{array}{c}x\approx 2,\!465 \end{array}}$ ________ •    $30^x=100$ $(3\cdot 10)^x=10^2$ Aplicando logaritmo aos dois lados da equação, $\mathrm{\ell og}\big[(3\cdot 10)^x\big]=\mathrm{\ell og}(10^2)\\\\ x\cdot \mathrm{\ell og}(3\cdot 10)=2\cdot \mathrm{\ell og}\,10\\\\ x\cdot \mathrm{\ell og}(3\cdot 10)=2\cdot 1\\\\ x=\dfrac{2}{\mathrm{\ell og}(3\cdot 10)}$ (o log do produto é a soma dos logs ) $x=\dfrac{2}{\mathrm{\ell og}\,3+\mathrm{\ell og}\,10}\\\\\\ x=\dfrac{2}{\mathrm{\ell og}\,3+1}$ Substituindo pelo valor aproximado, $x\approx \dfrac{2}{0,\!477+1}\\\\\\ \boxed{\begin{array}{c}x\approx 1,\!354 \end{array}}$ ________ •    $20^x=3^{x-2}$ $(2\cdot 10)^x=3^{x-2}$ Aplicando logaritmo aos dois lados $\mathrm{\ell og}\big[(2\cdot 10)^x\big]=\mathrm{\ell og}(3^{x-2})\\\\ x\,\mathrm{\ell og}(2\cdot 10)=(x-2)\cdot \mathrm{\ell og}\,3$ Aplicando a distributiva no lado direito, $x\,\mathrm{\ell og}(2\cdot 10)=x\,\mathrm{\ell og\,}3-2\,\mathrm{\ell og}\,3\\\\ x\,\mathrm{\ell og}(2\cdot 10)-x\,\mathrm{\ell og\,}3=-2\,\mathrm{\ell og}\,3\\\\ x\cdot \big[\mathrm{\ell og}(2\cdot 10)-\mathrm{\ell og\,}3\big]=-2\,\mathrm{\ell og}\,3\\\\ x\cdot \big[\mathrm{\ell og\,}2+\mathrm{\ell og\,}10-\mathrm{\ell og\,}3\big]=-2\,\mathrm{\ell og}\,3\\\\ x\cdot \big[\mathrm{\ell og\,}2+1-\mathrm{\ell og\,}3\big]=-2\,\mathrm{\ell og}\,3\\\\ x=\dfrac{-2\,\mathrm{\ell og\,}3}{\mathrm{\ell og\,}2+1-\mathrm{\ell og\,}3}$ Substituindo pelos valores aproximados, $x\approx \dfrac{-2\cdot 0,\!477}{0,\!301+1-0,\!477}\\\\\\ x\approx \dfrac{-0,\!954}{0,\!824}\\\\\\ \boxed{\begin{array}{c}x\approx -1,\!158 \end{array}}$ Dúvidas? Comente. Bons estudos! :-)