Respostas

2014-07-11T18:22:19-03:00
Olá Mila,

use a propriedade da potenciação, e use uma variável auxiliar (k):

3^{ x^{2} }-4*3^{x}+3=0\\
(3^x)^2-4*3^x+3=0\\\\
3^x=k\\\\
k^2-4k+3=0\\\\
\Delta=b^2-4ac\\
\Delta=(-4)^2-4*1*3\\
\Delta=16-12\\
\Delta=4\\\\
k= \dfrac{-b\pm \sqrt{\Delta} }{2}= \dfrac{-(-4)\pm \sqrt{4} }{2}= \dfrac{4\pm2}{2}\\\\\\
\begin{cases}k'= \dfrac{4-2}{2}= \dfrac{2}{2}=1\\\\
k''= \dfrac{4+2}{2}= \dfrac{6}{2}=3    \end{cases}

Retomando a variável original, usando os valores de k, temos os valores de x:

~~~~~~~~~~~~~~~~~~~~3^x=k'\\\\
3^x=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~3^x=3\\
3^x=3^0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~3^x=3^1\\
\not3^x=\not3^0~~~~~~~~~~~~~~~~~~~~~~~~~~~\not3^x=\not3^1\\\\
x'=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x''=1\\\\
\boxed{S=\{0,1\}}

Espero ter ajudado e tenha ótimos estudos =))