Question

a solução da equação exponencial. (4^x) +(6^x)=2.(9^x)

Answer 1

4ˣ + 6ˣ = 2.9ˣ

Dividimos a equação por 9ˣ:

4ˣ/9ˣ + 6ˣ/9ˣ = 2.9ˣ/9ˣ
(2/3)ˣ.(2/3)ˣ + (2/3)ˣ = 2

Seja y = (2/3)ˣ:
y² + y = 2
y² + y - 2 = 0
Δ = 1² - 4.1.(-2) = 1 + 8 = 9

y' = (-1 + √9) / (2.1) = (-1 + 3)/2 = 2/2 = 1
y'' = (-1 - √9) / (2.1) = (-1 - 3)/2 = -4/2 = -2

y = (2/3)ˣ => (2/3)ˣ = 1 => x = 0 ou (2/3)ˣ = -2 (impossível)

Portanto, x = 0.

Answer 2

Explicação passo-a-passo:

$\sf (4^x)+(6^x)=2\cdot(9^x)$

$\sf (2^2)^x+(2\cdot3)^x=2\cdot(3^2)^x$

$\sf 2^{2x}+2^x\cdot3^x=2\cdot(3^x)^2$

$\sf (2^{2x})+2^x\cdot3^x-2\cdot3^{2x}=0$

$\sf \dfrac{2^{2x}}{3^{2x}}+\dfrac{2^{x}\cdot3^{x}}{3^{2x}}-\dfrac{2\cdot3^{2x}}{3^{2x}}=0$

$\sf \left[\left(\dfrac{2}{3}\right)^x]^2+\left(\dfrac{2}{3}\right)^x-2=0$

Seja $\sf y=\left(\dfrac{2}{3}\right)^x$

$\sf y^2+y-2=0$

$\sf \Delta=1^2-4\cdot1\cdot(-2)$

$\sf \Delta=1+8$

$\sf \Delta=9$

$\sf y=\dfrac{-1\pm\sqrt{9}}{2\cdot1}=\dfrac{-1\pm3}{2}$

$\sf y'=\dfrac{-1+3}{2}~\Rightarrow~y'=\dfrac{2}{2}~\Rightarrow~y'=1$

$\sf y"=\dfrac{-1-3}{2}~\Rightarrow~y'=\dfrac{-4}{2}~\Rightarrow~\underbrace{\red{y"=-2}}_{não~serve,~pois~y >0}$

Assim:

$\sf \left(\dfrac{2}{3}\right)^x=1$

$\sf \left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^0$

Igualando os expoentes:

$\sf x=0$

$\sf S=\{0\}$