Matemática
tatyberilo
1

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

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

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