Question

resolva as equação exponenciais:

a) $( \frac{1}{3} ) ^{x+1} = \frac{ \sqrt{3} }{9}$

b)$\sqrt{ (\frac{1}{2})}^{3x-2} = (\frac{1}{2} ) ^{-4x} .2 ^{-x+4}$


c)$( \frac{1}{27} )^{-x} .(3 ^{3x}) ^2= (\frac{1}{3} ) ^{x-1}$

a numero b o 3x-2 esta dentro da raiz quadrada.

Answer 1

Olá Ilane,

use as propriedades da potenciação:

$\left( \dfrac{1}{3}\right)^{x+1}= \dfrac{ \sqrt{3} }{9}\\\\ (3^{-1})^{x+1}= \sqrt{3}*9^{-1}\\\\ 3^{-x-1}=3^{ \tfrac{1}{2} }*(3^2)^{-1}\\ 3^{-x-1}=3^{ \tfrac{1}{2} }*3^{-2}\\ 3^{-x-1}=3^{ \tfrac{1}{2}-2 }\\ 3^{-x-1}=3^{- \tfrac{3}{2} }\\ \not3^{-x-1}=\not3^{- \tfrac{3}{2} }\\\\ -x-1=- \dfrac{3}{2}\\\\ -x=- \dfrac{3}{2}+1\\\\ -x=- \dfrac{1}{2}~~*~~(-1)\\\\ x= \dfrac{1}{2}\\\\ S=\left\{ \dfrac{1}{2}\right\}$

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$\sqrt{\left( \dfrac{1}{2}\right)^{3x-2} }=\left( \dfrac{1}{2}\right)^{-4x}*2^{-x+4}\\\\ \left( \dfrac{1}{2}\right)^{ \tfrac{3x-2}{2} }=(2^{-1})^{-4x}*2^{-x+4}\\\\ (2^{-1})^{ \tfrac{3x-2}{2} }=2^{4x}*2^{-x+4}\\\\ \not2^{- \tfrac{3x-2}{2} }=\not2^{3x+4}\\\\ -\dfrac{3x-2}{2}=3x+4\\\\ -(3x-2)=2(3x+4)\\ -3x+2=6x+8\\ -3x-6x=8-2\\ -9x=6\\\\ x= \dfrac{~~6}{-9}\\\\ x=- \dfrac{2}{3}\\\\ S=\left\{- \dfrac{2}{3}\right\}$

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$\left( \dfrac{1}{27}\right)^{-x}*(3^{3x})^2=\left( \dfrac{1}{3}\right)^{x-1}\\\\ (3^{-3})^{-x}*3^{6x}=(3^{-1})^{x-1}\\\\ 3^{3x}*3^{6x}=3^{-x+1}\\\\ \not3^{9x}=\not3^{-x+1}\\\\ 9x=-x+1\\ 9x+x=1\\ 10x=1\\\\ x= \dfrac{1}{10}\\\\ S=\left\{ \dfrac{1}{10}\right\}$

Espero ter ajudado e tenha ótimos estudos =))