Question

Resolva a equação exponencial 3^2x+1 - 12. 3* + 9 = 0

Answer 1

Resposta:

Solução:

$\sf \displaystyle 3^{2x+1} - 12\cdot 3x + 9 = 0$

$\sf \displaystyle 3^{2x} \cdot 3^1 - 12\cdot 3^x + 9 = 0$

$\sf \displaystyle 3\cdot (3^x)^2 - 12\cdot 3^x + 9 = 0$

Fazendo $\sf \textstyle 3^x = y$, temos:

$\sf \displaystyle 3y^2- 12y +9 = 0$

$\sf \displaystyle \Delta = b^2 -\:4ac$

$\sf \displaystyle \Delta = (-12)^2 -\:4 \cdot 3 \cdot 9$

$\sf \displaystyle \Delta = 144 -108$

$\sf \displaystyle \Delta = 36$

$\sf \displaystyle y = \dfrac{-\,b \pm \sqrt{ \Delta } }{2a}$

$\sf \displaystyle y = \dfrac{-\,(-12) \pm \sqrt{36 } }{2 \cdot 3}$

$\sf \displaystyle y = \dfrac{12 \pm 6}{6}$

$\sf \begin{cases} \sf y_1 = &\sf \dfrac{12+ 6}{6} = \dfrac{18}{6} = 3 \\\\ \sf y_2 = &\sf \dfrac{12 - 6}{6} = \dfrac{6}{6} = 1\end{cases}$

Como $\sf \textstyle 3^x = y$, temos:

$\sf \displaystyle 3^x = y_1$

$\sf \displaystyle 3^x = 3$

$\sf \displaystyle \diagup\!\!\!{ 3}^x = \diagup\!\!\!{ 3}^1$

$\boxed{ \boxed { \boldsymbol{ \sf \displaystyle x = 1}}} \quad \gets \mathbf{ Resposta }$

$\sf \displaystyle 3^x = y_2$

$\sf \displaystyle 3^x = 1$

$\sf \displaystyle \diagup\!\!\!{ 3}^x = \diagup\!\!\!{ 3}^0$

$\boxed{ \boxed { \boldsymbol{ \sf \displaystyle x = 0}}} \quad \gets \mathbf{ Resposta }$

Logo, S = { 0, 1 }.

Explicação passo-a-passo:

Answer 2

Resposta:

$\textsf{Leia abaixo}$

Explicação passo-a-passo:

$\mathsf{3^{2x + 1} - 12.3^x + 9 = 0}$

$\mathsf{3.(3^x)^2 - 12.(3^x) + 9 = 0}$

$\mathsf{3^x = y}$

$\mathsf{3y^2 - 12y + 9 = 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = (-12)^2 - 4.3.9}$

$\mathsf{\Delta = 144 - 108}$

$\mathsf{\Delta = 36}$

$\mathsf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{12 \pm \sqrt{36}}{6} \rightarrow \begin{cases}\mathsf{y' = \dfrac{12 + 6}{6} = \dfrac{18}{6} = 3}\\\\\mathsf{y'' = \dfrac{12 - 6}{6} = \dfrac{6}{6} = 1}\end{cases}}$

$\mathsf{3^x = y'}$

$\mathsf{\not3^x = \not3^1}$

$\mathsf{x = 1}$

$\mathsf{3^x = y''}$

$\mathsf{3^x = 1}$

$\mathsf{\not3^x = \not3^0}$

$\mathsf{x = 0}$

$\boxed{\boxed{\mathsf{S = \{1;0\}}}}$