Question

Resolva as seguintes equações exponenciais:

 a) $81^{x-2}= \sqrt[4]{27}$

b) $\sqrt{ 4^{x+1} } = \sqrt[3]{16}$

c)$\sqrt{ 5^{x} } . 25^{x+1} = (0,2)^{1-x}$

d) $( \frac{2}{5}) ^{x+3} = (\frac{125}{8} ) ^{x-1} .(0,4)^{2x-3}$

e)$\sqrt[5]{2^x}. \sqrt[3]{4^x} = \sqrt{8 ^{-x} }$

Answer 1

Olá Ilane,

Use as propriedades da potenciação e transforme os números decimais em fração:

$81^{x-2}= \sqrt[4]{27}\\ (3^4)^{x-2}= \sqrt[4]{3^3}\\ 3^{4x-8}=3^{ \tfrac{3}{4} }\\\\ \not3^{4x-8}=\not3^{ \tfrac{3}{4} }\\\\ 4x-8= \dfrac{3}{4}\\\\ 4x= \dfrac{3}{4}+8\\\\ 4x= \dfrac{35}{4}\\\\ x= \dfrac{35}{12}\\\\S=\left\{ \dfrac{35}{12}\right\}$

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$\sqrt{4^{x+1} }= \sqrt[3]{16}\\\\ 4^{ \tfrac{x+1}{2} }= \sqrt[3]{4^2}\\\\ 4^{ \tfrac{x+1}{2} }=4^{ \tfrac{2}{3} }\\\\ \not4^{ \tfrac{x+1}{2} }=\not4^{ \tfrac{2}{3} }\\\\ \dfrac{x+1}{2}= \dfrac{2}{3}\\\\ 3(x+1)=2*2\\\\ 3x+3=4\\\\ 3x=4-3\\\\ 3x=1\\\\ x= \dfrac{1}{3}\\\\ S=\left\{ \dfrac{1}{3}\right\}$

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$\sqrt{5^x}*25^{x+1}=(0,2)^{1-x}\\\\ 5^{ \tfrac{x}{2} }*(5^2)^{x+1}= \left(\dfrac{1}{5}\right)^{1-x}\\\\\\ 5^{ \tfrac{x}{2} }*5^{2x+2}=(5^{-1})^{1-x}\\\\ 5^{ \tfrac{x}{2}+2x+2 }=5^{x-1}\\\\ \not5^{ \tfrac{x}{2}+2x+2 }=\not5^{x-1}\\\\ \dfrac{x}{2}+2x+2=x-1\\\\ \dfrac{x}{2}=x-1-2x-2\\\\ \dfrac{x}{2}=-x-3\\\\ x=2(-x-3)\\ x=-2x-6\\ x+2x=-6\\ 3x=-6\\\\ x= \dfrac{-6}{~3}\\\\ x=-2\\\\ S=\{-2\}$

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$\left(\dfrac{2}{5}\right)^{x+3}= \left(\dfrac{125}{8}\right)^{x-1}*(0,4)^{2x-3}\\\\ \left(\dfrac{2}{5}\right)^{x+3}=\left( \dfrac{5^3}{2^3}\right)^{x-1}*\left( \dfrac{2}{5}\right)^{2x-3}\\\\ \left( \dfrac{2}{5}\right)^{x+3}=\left( \dfrac{5}{2}\right)^{3(x-1)}*\left( \dfrac{2}{5}\right) ^{2x-3}\\\\ \left( \dfrac{2}{5}\right)^{x+3}=\left( \dfrac{2}{5}\right)^{-3*(x-1)}*\left( \dfrac{2}{5}\right)^{2x-3}$

$\left( \dfrac{2}{5}\right)^{x+3}=\left( \dfrac{2}{5}\right)^{-3x+3}*\left( \dfrac{2}{5}\right)^{2x-3}\\\\ \left( \dfrac{2}{5}\right)^{x+3}=\left( \dfrac{2}{5}\right)^{-3x+3+2x-3}\\\\ \left( \dfrac{2}{5}\right)^{x+3}=\left( \dfrac{2}{5}\right)^{-x}\\\\\\ x+3=-x\\ -x-x=3\\ -2x=3\\\\ x= \dfrac{~~3}{-2}\\\\ x=- \dfrac{3}{2}\\\\ S=\left\{- \dfrac{3}{2}\right\}$

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$\sqrt[5]{2^x}* \sqrt[3]{4^x}= \sqrt{8^{-x} }\\\\ 2^{ \tfrac{x}{5} }* \sqrt[3]{(2^2)^{x} }= \sqrt{(2^3)^{-x} }\\\\ 2^{ \tfrac{x}{5} }* \sqrt[3]{2 ^{2x}}= \sqrt{2^{-3x} }\\\\ 2^{ \tfrac{x}{5} }*2^{ \tfrac{2}{3}x }=2^{-3x}\\\\ \not2^{ \tfrac{x}{5}+ \tfrac{2}{3}x}=\not2^{-3x}\\\\ \dfrac{x}{5}+ \dfrac{2}{3}x=-3x$

$\dfrac{13x}{15}=-3x\\\\ 13x=15*(-3x)\\\\ 13x=-45x~\to~sem~soluc\~ao$

Espero ter ajudado e tenha ótimos estudos =))