1. **State the problem:** Solve the equation $$2^x - 2^{-x} = 1$$ for $x$. 2. **Rewrite the equation:** Recall that $2^{-x} = \frac{1}{2^x}$. So the equation becomes: $$2^x - \frac{1}{2^x} = 1$$ 3. **Substitute:** Let $y = 2^x$. Since $2^x > 0$ for all real $x$, $y > 0$. The equation becomes: $$y - \frac{1}{y} = 1$$ 4. **Multiply both sides by $y$ to clear the denominator:** $$y^2 - 1 = y$$ 5. **Rearrange into a quadratic equation:** $$y^2 - y - 1 = 0$$ 6. **Solve the quadratic equation:** Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-1$, $c=-1$: $$y = \frac{1 \pm \sqrt{1^2 - 4 \times 1 \times (-1)}}{2} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$$ 7. **Select the positive root:** Since $y = 2^x > 0$, we take the positive root: $$y = \frac{1 + \sqrt{5}}{2}$$ 8. **Solve for $x$:** $$2^x = \frac{1 + \sqrt{5}}{2}$$ Taking the logarithm base 2 on both sides: $$x = \log_2 \left( \frac{1 + \sqrt{5}}{2} \right)$$ **Final answer:** $$x = \log_2 \left( \frac{1 + \sqrt{5}}{2} \right)$$