1. **State the problem:** Solve the inequality $$2^x + 2 \cdot 2^{-x} - 3 > 0$$. 2. **Rewrite the terms:** Let $$y = 2^x$$. Then $$2^{-x} = \frac{1}{2^x} = \frac{1}{y}$$. 3. **Substitute into the inequality:** $$y + 2 \cdot \frac{1}{y} - 3 > 0$$ 4. **Multiply both sides by $$y$$ (note $$y > 0$$ since $$2^x > 0$$):** $$y \cdot y + y \cdot 2 \cdot \frac{1}{y} - y \cdot 3 > 0 \implies y^2 + 2 - 3y > 0$$ 5. **Rewrite the inequality:** $$y^2 - 3y + 2 > 0$$ 6. **Factor the quadratic:** $$y^2 - 3y + 2 = (y - 1)(y - 2)$$ 7. **Analyze the inequality:** $$(y - 1)(y - 2) > 0$$ means the product is positive. This happens when both factors are positive or both are negative. - Case 1: $$y - 1 > 0$$ and $$y - 2 > 0 \implies y > 2$$ - Case 2: $$y - 1 < 0$$ and $$y - 2 < 0 \implies y < 1$$ 8. **Recall $$y = 2^x$$, which is always positive. So the solution in terms of $$x$$ is:** $$2^x < 1 \quad \text{or} \quad 2^x > 2$$ 9. **Solve each inequality:** - $$2^x < 1 \implies x < 0$$ - $$2^x > 2 \implies x > 1$$ 10. **Final solution:** $$x < 0 \quad \text{or} \quad x > 1$$ This means the inequality holds for all $$x$$ less than 0 or greater than 1.