1. **State the problem:** Solve the equation $$2^x + 3^x - 4^x + 6^x - 9^x = 1.$$\n\n2. **Rewrite terms using powers:** Notice that $$4^x = (2^2)^x = 2^{2x}$$ and $$9^x = (3^2)^x = 3^{2x}.$$\n\n3. **Substitute:** The equation becomes $$2^x + 3^x - 2^{2x} + 6^x - 3^{2x} = 1.$$\n\n4. **Express $$6^x$$ as $$2^x \cdot 3^x$$:** So the equation is $$2^x + 3^x - 2^{2x} + 2^x 3^x - 3^{2x} = 1.$$\n\n5. **Let $$a = 2^x$$ and $$b = 3^x$$:** Then the equation is $$a + b - a^2 + ab - b^2 = 1.$$\n\n6. **Rewrite the equation:** $$a + b + ab - a^2 - b^2 = 1.$$\n\n7. **Group terms:** $$-a^2 + ab - b^2 + a + b = 1.$$\n\n8. **Try to factor or rearrange:** Note that $$-a^2 - b^2 + ab = - (a^2 - ab + b^2).$$\n\n9. **Recall that $$a^2 - ab + b^2 = (a - \frac{b}{2})^2 + \frac{3b^2}{4} > 0$$ for all real $$a,b$$ except zero. So the expression is negative definite.\n\n10. **Check simple values:** Try $$x=0$$: $$2^0 + 3^0 - 4^0 + 6^0 - 9^0 = 1 + 1 - 1 + 1 - 1 = 1.$$ So $$x=0$$ is a solution.\n\n11. **Check $$x=1$$:** $$2 + 3 - 4 + 6 - 9 = (5 - 4) + (6 - 9) = 1 - 3 = -2 \neq 1.$$\n\n12. **Check $$x=2$$:** $$4 + 9 - 16 + 36 - 81 = 13 - 16 + 36 - 81 = -3 + 36 - 81 = 33 - 81 = -48 \neq 1.$$\n\n13. **Check $$x=-1$$:** $$\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{6} - \frac{1}{9} = 0.5 + 0.3333 - 0.25 + 0.1667 - 0.1111 = 0.639 \neq 1.$$\n\n14. **Conclusion:** The only solution is $$x=0$$.