Let \(u = 4^x\). Then \(4^{-x} = \frac{1}{u}\).

The equation becomes \(u = 3 + \frac{1}{u}\).

Multiply through by \(u\) to clear the fraction: \(u^2 = 3u + 1\).

Rearrange to form a quadratic equation: \(u^2 - 3u - 1 = 0\).

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -3\), \(c = -1\).

\(u = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}\).

Since \(u = 4^x\) must be positive, take the positive root: \(u = \frac{3 + \sqrt{13}}{2}\).

Calculate \(u \approx 3.30277563773\).

Since \(u = 4^x\), solve for \(x\): \(x = \log_4(u)\).

\(x = \frac{\log(u)}{\log(4)}\).

\(x \approx \frac{\log(3.30277563773)}{\log(4)} \approx 0.862\).