Question

preciso de ajuda nessas questões de equação exponencial


${25}^{3x + 2} = \frac{1}{5 { }^{ - 4x } }$


${4}^{3x} = \sqrt[3]{32}$


${9}^{2x - 1} = \frac{1}{ \sqrt[4]{27} }$

Answer 1

$a)\\\\ 25^{3x+2} = \dfrac{1}{5^{-4x}}\\\\ 5^{2(3x+2)} = \dfrac{1}{5^{-4x}}\\\\ 5^{6x+4} = \dfrac{1}{5^{-4x}}\\\\ 5^{6x+4}\ .\ 5^{-4x}= 1\\\\ 5^{6x + 4 + (-4x)} = 1\\\\ 5^{2x+4}=5^0\\\\ 2x+4=0\\\\ x=\dfrac{-4}{2}\\\\ x=-2$

$b)\\\\ 4^{3x}= \sqrt[3]{32} \\\\ 2^{6x}= \sqrt[3]{2^5} \\\\ 2^{6x} = 2^{\frac{5}{3}}\\\\ 6x = \dfrac{5}{3}\\\\ x = \dfrac{5}{18}$

$c)\\\\ 9^{2x-1} = \dfrac{1}{ \sqrt[4]{27} }\\\\ 3^{4x-2} = \dfrac{3^0}{3^{\frac{3}{4}}} }\\\\ 3^{4x-2+\frac{3}{4}} = 3^0\\\\ 4x-2+\dfrac{3}{4} = 0\\\\ \dfrac{16x-8+3=0}{4}\\\\ 16x-8+3=0\\\\ 16x - 5 = 0\\\\ x = \dfrac{5}{16}$

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