Question

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

Answer 1

•    $3^{x-1}=5$


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! :-)