Question

Equação exponencial:

$2^{2x+2} - 6^{x} - 2\cdot3^{2x+2}=0$

Answer 1

$2^{2x+2}-6^{x}-2\cdot3^{2x+2}=0$

Dividindo todos os membros da igualdade por $3^{2x+2}$ (funcionaria também dividir por $2^{2x+2}$):

$\dfrac{2^{2x+2}}{3^{2x+2}}-\dfrac{6^{x}}{3^{2x+2}}-2\cdot\dfrac{3^{2x+2}}{3^{2x+2}}=0\\\\\\\bigg(\dfrac{2}{3}\bigg)^{2x+2}-\dfrac{(2\cdot3)^{x}}{3^{2x+2}}-2=0\\\\\\\bigg(\dfrac{2}{3}\bigg)^{2}\bigg(\dfrac{2}{3}\bigg)^{2x}-\dfrac{2^{x}\cdot3^{x}}{3^{2x+2}}-2=0\\\\\\\dfrac{4}{9}\bigg[\bigg(\dfrac{2}{3}\bigg)^{x}\bigg]^{2}-\dfrac{2^{x}}{3^{x+2}}-2=0\\\\\\\dfrac{4}{9}\bigg[\bigg(\dfrac{2}{3}\bigg)^{x}\bigg]^{2}-\dfrac{2^{x}}{3^{2}\cdot3^{x}}-2=0$

$\dfrac{4}{9}\bigg[\bigg(\dfrac{2}{3}\bigg)^{x}\bigg]^{2}-\dfrac{1}{9}\cdot\dfrac{2^{x}}{3^{x}}-2=0\\\\\\\dfrac{4}{9}\bigg[\bigg(\dfrac{2}{3}\bigg)^{x}\bigg]^{2}-\dfrac{1}{9}\bigg(\dfrac{2}{3}\bigg)^{x}-2=0$

Fazendo $y=\big(\frac{2}{3}\big)^{x}$, podemos resolver a equação em y

$\dfrac{4}{9}y^{2}-\dfrac{1}{9}y-2=0$

Multiplicando todos os membros por 9:

$4y^{2}-y-18=0$

Resolvemos essa equação por Bhaskara

$\Delta=b^{2}-4ac=(-1)^{2}-4\cdot4\cdot(-18)=1+288=289\\\\\\y=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-(-1)\pm\sqrt{289}}{2\cdot4}=\dfrac{1\pm17}{8}\begin{cases}y_{1}=\dfrac{1+17}{8}=\dfrac{9}{4}\\\\\\y_{2}=\dfrac{1-17}{8}=-2\end{cases}$

Como $y=\big(\frac{2}{3}\big)^{x}\ \textgreater \ 0$, descartamos o valor negativo encontrado para y

Portanto, encontramos y = 4/9. Podemos agora encontrar x:

$y=\dfrac{9}{4}\\\\\\\bigg(\dfrac{2}{3}\bigg)^{x}=\dfrac{9}{4}\\\\\\\bigg(\dfrac{2}{3}\bigg)^{x}=\dfrac{3^{2}}{2^{2}}\\\\\\\bigg(\dfrac{2}{3}\bigg)^{x}=\bigg(\dfrac{3}{2}\bigg)^{2}\\\\\\\bigg(\dfrac{2}{3}\bigg)^{x}=\bigg(\dfrac{2}{3}\bigg)^{-2}$

Então, a solução (real) da equação é x = - 2

Answer 2

Resposta:

2^{2x+2} - 6^{x} - 2 • 3^{2x+2} = 0

2^{2x} • 2^{2} - (2 • 3)^{x} - 2 • 3^{2x} • 3^{2} = 0

2^{2} • 2^{2x} - 2^{x} • 3^{x} - 2 • 3^{2} • 3^{2x} = 0

4 • 2^{2x} - 2^{x} • 3^{x} - 18 • 3^{2x} = 0 ⬅ Divida tudo por 2.

4 • (2/3)^{2x} - (2/3)^{x} - 18 = 0

4 • ((2/3)^{x})^{2} - (2/3)^{x} - 18 = 0 ⬅ t = (2/3)^{x}.

4t^{2} - t - 18 = 0

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↔Função Quadrática↔

ax² + bx + c = 0 ➡ x = -b±√{b² - 4ac}/2a

t = -(- 1)±√{(- 1)² - 4 • 4 • (- 18)}/2•4

t = 1±√{1 + 288}/8

t = 1±√289/8

t = 1±17/8

t = 1+17/8 = 9/4

t = 1-17/8 = - 2

==================================

t = 9/4

t = - 2

⬆t = (2/3)^{x}

(2/3)^{x} = 9/4 ➡ x = - 2

(2/3)^{x} = - 2 ➡ x∈∅

x = - 2