Start with the equation \(e^x = 1 + 6e^{-x}\).

Multiply both sides by \(e^x\) to eliminate the negative exponent: \(e^{2x} = e^x + 6\).

Rearrange to form a quadratic equation: \(e^{2x} - e^x - 6 = 0\).

Let \(u = e^x\), then the equation becomes \(u^2 - u - 6 = 0\).

Factor the quadratic: \((u - 3)(u + 2) = 0\).

Thus, \(u = 3\) or \(u = -2\). Since \(u = e^x\) and \(e^x > 0\), \(u = -2\) is not valid.

Therefore, \(e^x = 3\).

Take the natural logarithm of both sides: \(x = \\ln(3)\).

Calculate \(x\) to 3 significant figures: \(x = 1.10\).