1. **State the problem:** Solve the equation $$3^x + 9^x + 27^x = 14$$ for the variable $x$. 2. **Rewrite the terms with the same base:** Note that $9 = 3^2$ and $27 = 3^3$, so rewrite the equation as: $$3^x + (3^2)^x + (3^3)^x = 14$$ which simplifies to: $$3^x + 3^{2x} + 3^{3x} = 14$$ 3. **Substitute:** Let $$y = 3^x$$. Then the equation becomes: $$y + y^2 + y^3 = 14$$ 4. **Rewrite as a polynomial:** $$y^3 + y^2 + y - 14 = 0$$ 5. **Solve the cubic equation:** Try possible rational roots using factors of 14: $\pm1, \pm2, \pm7, \pm14$. Check $y=2$: $$2^3 + 2^2 + 2 = 8 + 4 + 2 = 14$$ which satisfies the equation. 6. **Factor the cubic:** Since $y=2$ is a root, divide the cubic by $(y-2)$: $$y^3 + y^2 + y - 14 = (y-2)(y^2 + 3y + 7)$$ 7. **Solve the quadratic:** $$y^2 + 3y + 7 = 0$$ Calculate discriminant: $$\Delta = 3^2 - 4 \times 1 \times 7 = 9 - 28 = -19 < 0$$ No real roots here. 8. **Real solution for $y$ is $2$ only.** 9. **Back-substitute for $x$:** $$3^x = 2$$ Take logarithm base 3: $$x = \log_3 2 = \frac{\ln 2}{\ln 3}$$ **Final answer:** $$x = \frac{\ln 2}{\ln 3}$$