1. **State the problem:** Solve the equation $$3^x = 2 + 3^{-x}$$ correct to 3 significant figures. 2. **Rewrite the equation:** Let $$u = 3^x$$. Then $$3^{-x} = \frac{1}{3^x} = \frac{1}{u}$$. 3. **Substitute into the equation:** $$u = 2 + \frac{1}{u}$$ 4. **Multiply both sides by $$u$$ to clear the denominator:** $$u^2 = 2u + 1$$ 5. **Rearrange to form a quadratic equation:** $$u^2 - 2u - 1 = 0$$ 6. **Solve the quadratic equation using the quadratic formula:** $$u = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times (-1)}}{2} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$$ 7. **Evaluate the two possible values:** - $$u = 1 + \sqrt{2} \approx 1 + 1.414 = 2.414$$ - $$u = 1 - \sqrt{2} \approx 1 - 1.414 = -0.414$$ (discard since $$u = 3^x > 0$$) 8. **Find $$x$$ from $$u = 3^x$$:** $$3^x = 2.414 \implies x = \log_3(2.414) = \frac{\ln(2.414)}{\ln(3)}$$ 9. **Calculate $$x$$:** $$x \approx \frac{0.881}{1.099} = 0.802$$ **Final answer:** $$x \approx 0.802$$ (3 significant figures)