Start by substituting \(u = 3^x\). Then \(3^{-x} = \frac{1}{3^x} = \frac{1}{u}\).

The equation becomes \(u = 2 + \frac{1}{u}\).

Multiply through by \(u\) to clear the fraction: \(u^2 = 2u + 1\).

Rearrange to form a quadratic equation: \(u^2 - 2u - 1 = 0\).

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1, b = -2, c = -1\).

Calculate the discriminant: \(b^2 - 4ac = (-2)^2 - 4(1)(-1) = 4 + 4 = 8\).

Thus, \(u = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}\).

Since \(u = 3^x\) must be positive, choose \(u = 1 + \sqrt{2}\).

Then \(3^x = 1 + \sqrt{2}\).

Take the logarithm of both sides: \(x \log 3 = \log(1 + \sqrt{2})\).

Thus, \(x = \frac{\log(1 + \sqrt{2})}{\log 3}\).

Calculate \(x \approx 0.802\) to 3 significant figures.