1/4·(3^x + 3^(x + 1)) = √3^(1 - x)
pongo:
3^x = t > 0
1/4·(t + 3·t) = √(3/t)
t = √(3/t)
elevo al quadrato:
t^2 = 3/t----> t = 3^(1/3)
3^x = 3^(1/3)-----> x = 1/3
$ \frac{1}{4} (3^x +3\cdot 3^x) = \frac{3^{\frac{1}{2}}} {\frac{x}{2}} $
$ \frac{1}{4} 3^x (1+3) = \frac{3^{\frac{1}{2}}}{3^\frac{x}{2}} $
$ 3^x = \frac{3^{\frac{1}{2}}}{3^\frac{x}{2}} $
$ 3^{x(1+\frac{1}{2})} = 3^{\frac{1}{2}} $
$ \frac{3}{2}x = \frac{1}{2} \; ⇒ \; x = \frac{1}{3} $