1. **State the problem:** Solve the equation $$\frac{9^x + 27^x}{9^x} = 10$$ for $x$. 2. **Rewrite the equation:** We can split the fraction: $$\frac{9^x}{9^x} + \frac{27^x}{9^x} = 10$$ which simplifies to $$1 + \frac{27^x}{9^x} = 10$$ 3. **Express bases as powers of 3:** Note that $9 = 3^2$ and $27 = 3^3$, so $$9^x = (3^2)^x = 3^{2x}$$ $$27^x = (3^3)^x = 3^{3x}$$ 4. **Substitute back:** $$1 + \frac{3^{3x}}{3^{2x}} = 10$$ 5. **Simplify the fraction:** $$\frac{3^{3x}}{3^{2x}} = 3^{3x - 2x} = 3^x$$ 6. **Rewrite the equation:** $$1 + 3^x = 10$$ 7. **Isolate $3^x$:** $$3^x = 10 - 1 = 9$$ 8. **Solve for $x$:** Since $9 = 3^2$, we have $$3^x = 3^2 \implies x = 2$$ **Final answer:** $$x = 2$$