1. **State the problem:** Solve the equation $$4^x + 6^x = 9^x$$ for $x$. 2. **Rewrite the equation:** Notice that $4 = 2^2$, $6 = 2 \cdot 3$, and $9 = 3^2$. However, it is simpler to work directly with the given bases. 3. **Divide both sides by $9^x$ to normalize:** $$\frac{4^x}{9^x} + \frac{6^x}{9^x} = 1$$ which simplifies to $$\left(\frac{4}{9}\right)^x + \left(\frac{6}{9}\right)^x = 1$$ 4. **Simplify the fractions:** $$\left(\frac{4}{9}\right)^x + \left(\frac{2}{3}\right)^x = 1$$ 5. **Set $y = \left(\frac{2}{3}\right)^x$:** Note that $\left(\frac{4}{9}\right)^x = \left(\frac{2}{3}\right)^{2x} = y^2$. 6. **Rewrite the equation in terms of $y$:** $$y^2 + y = 1$$ 7. **Solve the quadratic equation:** $$y^2 + y - 1 = 0$$ Using the quadratic formula: $$y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$$ 8. **Evaluate the roots:** $$y_1 = \frac{-1 + \sqrt{5}}{2} \approx 0.618$$ $$y_2 = \frac{-1 - \sqrt{5}}{2} \approx -1.618$$ Since $y = \left(\frac{2}{3}\right)^x > 0$, discard $y_2$. 9. **Solve for $x$ using $y_1$:** $$\left(\frac{2}{3}\right)^x = 0.618$$ Take natural logarithm on both sides: $$x \ln\left(\frac{2}{3}\right) = \ln(0.618)$$ 10. **Calculate $x$:** $$x = \frac{\ln(0.618)}{\ln\left(\frac{2}{3}\right)}$$ 11. **Approximate values:** $$\ln(0.618) \approx -0.48$$ $$\ln\left(\frac{2}{3}\right) \approx -0.405$$ 12. **Final answer:** $$x \approx \frac{-0.48}{-0.405} \approx 1.185$$ **Therefore, the solution to the equation is approximately** $x \approx 1.185$.