1. **State the problem:** Solve for all real values of $x$ in the equation $$8^x + 2^x = 130.$$\n\n2. **Rewrite the bases:** Note that $8 = 2^3$, so we can write $$8^x = (2^3)^x = 2^{3x}.$$\n\n3. **Substitute:** The equation becomes $$2^{3x} + 2^x = 130.$$\n\n4. **Use substitution:** Let $$y = 2^x.$$ Since $2^x > 0$ for all real $x$, $y > 0$. The equation becomes $$y^3 + y = 130.$$\n\n5. **Rewrite the equation:** $$y^3 + y - 130 = 0.$$\n\n6. **Solve the cubic equation:** We look for integer roots by testing factors of 130. Try $y=5$: $$5^3 + 5 = 125 + 5 = 130,$$ which satisfies the equation.\n\n7. **Factor the cubic:** Since $y=5$ is a root, divide by $(y-5)$: $$y^3 + y - 130 = (y - 5)(y^2 + 5y + 26).$$\n\n8. **Solve quadratic:** $$y^2 + 5y + 26 = 0.$$ The discriminant is $$\Delta = 5^2 - 4 \times 1 \times 26 = 25 - 104 = -79 < 0,$$ so no real roots here.\n\n9. **Real solution for $y$:** Only $y=5$ is real and positive.\n\n10. **Back-substitute:** Recall $y = 2^x$, so $$2^x = 5.$$\n\n11. **Solve for $x$:** Take logarithm base 2: $$x = \log_2 5 = \frac{\ln 5}{\ln 2}.$$\n\n**Final answer:** $$x = \log_2 5.$$