Question

3^2^×+2.3x-15 = 0

help???

Answer 1

$\large\begin{array}{l} \textsf{Resolver a equa\c{c}\~ao exponencial:}\\\\ \mathsf{3^{2x}+2\cdot 3x-15=0}\\\\ \mathsf{3^{x\,\cdot\,2}+2\cdot 3x-15=0}\\\\ \mathsf{(3^x)^2+2\cdot 3x-15=0} \end{array}$


$\large\begin{array}{l} \textsf{Fa\c{c}a a seguinte mudan\c{c}a de vari\'avel:}\\\\ \mathsf{3^{x}=t\qquad(t>0)}\\\\\\ \textsf{e a equa\c{c}\~ao fica}\\\\ \mathsf{t^2+2t-15=0}\quad\Rightarrow\quad\left\{\! \begin{array}{l} \mathsf{a=1}\\\mathsf{b=2}\\\mathsf{c=-15} \end{array} \right. \end{array}$


$\large\begin{array}{l} \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=2^2-4\cdot 1\cdot (-15)}\\\\ \mathsf{\Delta=4+60}\\\\ \mathsf{\Delta=64} \end{array}$


$\large\begin{array}{l} \mathsf{t=\dfrac{-b\pm \sqrt{\Delta}}{2a}}\\\\ \mathsf{t=\dfrac{-2\pm \sqrt{64}}{2\cdot 1}}\\\\ \mathsf{t=\dfrac{-2\pm 8}{2}}\\\\ \mathsf{t=\dfrac{\diagup\!\!\!\! 2\cdot (-1\pm 4)}{\diagup\!\!\!\! 2}} \end{array}$


$\large\begin{array}{l} \mathsf{t=-1\pm 4}\\\\ \begin{array}{rcl} \mathsf{t=-1+4}&~\textsf{ ou }~&\mathsf{t=-1-4}\\\\ \mathsf{t=3}&~\textsf{ ou }~&\mathsf{t=-5}\quad\textsf{(n\~ao serve, pois }\mathsf{-5<0}\textsf{)} \end{array} \end{array}$


$\large\begin{array}{l} \textsf{Ent\~ao temos}\\\\ \mathsf{t=3}\\\\\\ \textsf{Voltando \'a vari\'avel x, ficamos com}\\\\ \mathsf{3^x=3} \\\\ \mathsf{3^x=3^1} \end{array}$


$\large\begin{array}{l} \textsf{Temos acima uma igualdade entre exponenciais de mesma}\\\textsf{base. Ent\~ao, \'e s\'o igualar os expoentes:}\\\\ \boxed{\begin{array}{c}\mathsf{x=1} \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a solu\c{c}\~ao.}\\\\\\ \textsf{Conjunto solu\c{c}\~ao: }\mathsf{S=\{1\}.} \end{array}$


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$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$


Tags: equação exponencial mudança de variável substituição quadrática segundo grau solução resolver álgebra