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	<title>Comments on: Equação exponencial</title>
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	<description>Exercícios para o Vestibular</description>
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		<title>By: thiago</title>
		<link>http://www.amigonerd.com/equacao-exponencial/comment-page-1/#comment-53331</link>
		<dc:creator>thiago</dc:creator>
		<pubDate>Fri, 09 Dec 2011 23:15:53 +0000</pubDate>
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		<description>8^x - 3*4^x - 3*2^(x+1) + 8 = 0
(2^x)^3 - 3*(2^x)^2 - 3*2^1*2^x + 2^3 = 0
2^x = y
y^3 - 3*y^2 - 6y + 8 = 0
como a soma dos coeficientes &#233; zero, uma das raizes &#233; 1
por briot ruffini, divide-se o polinomio por (y - 1)
&#160;&#160;&#160; &#124;&#160; 1&#160;&#160; -3&#160;&#160;&#160; -6&#160;&#160; 8
1&#160; &#124;&#160; 1&#160;&#160; -2&#160;&#160;&#160; -8&#160;&#160; 0
y^2 - 2*y - 8 = 0
ra&#237;zes 4 e -2
as ra&#237;zes s&#227;o {-2, 1, 4}
lembrando que 2^x = y; os resultado negativo &#233; impossivel, para os demais:
2^x = 1
x = 0
2^x = 4
x = 2
logo, o conjunto solu&#231;&#227;o:&#160;S = {0; 2}</description>
		<content:encoded><![CDATA[<p>8^x &#8211; 3*4^x &#8211; 3*2^(x+1) + 8 = 0<br />
(2^x)^3 &#8211; 3*(2^x)^2 &#8211; 3*2^1*2^x + 2^3 = 0<br />
2^x = y<br />
y^3 &#8211; 3*y^2 &#8211; 6y + 8 = 0<br />
como a soma dos coeficientes &eacute; zero, uma das raizes &eacute; 1<br />
por briot ruffini, divide-se o polinomio por (y &#8211; 1)<br />
&nbsp;&nbsp;&nbsp; |&nbsp; 1&nbsp;&nbsp; -3&nbsp;&nbsp;&nbsp; -6&nbsp;&nbsp; 8<br />
1&nbsp; |&nbsp; 1&nbsp;&nbsp; -2&nbsp;&nbsp;&nbsp; -8&nbsp;&nbsp; 0<br />
y^2 &#8211; 2*y &#8211; 8 = 0<br />
ra&iacute;zes 4 e -2<br />
as ra&iacute;zes s&atilde;o {-2, 1, 4}<br />
lembrando que 2^x = y; os resultado negativo &eacute; impossivel, para os demais:<br />
2^x = 1<br />
x = 0<br />
2^x = 4<br />
x = 2<br />
logo, o conjunto solu&ccedil;&atilde;o:&nbsp;S = {0; 2}</p>
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