1. **State the problem:** Solve the equation $$3 \cdot 4^x + 6 \cdot 2^x = 24$$ for $x$. 2. **Recall the formula and rules:** Note that $4^x = (2^2)^x = 2^{2x}$. This allows us to express all terms with the same base, which simplifies solving. 3. **Rewrite the equation:** Substitute $4^x$ with $2^{2x}$: $$3 \cdot 2^{2x} + 6 \cdot 2^x = 24$$ 4. **Use substitution:** Let $y = 2^x$. Then $2^{2x} = (2^x)^2 = y^2$. The equation becomes: $$3y^2 + 6y = 24$$ 5. **Bring all terms to one side:** $$3y^2 + 6y - 24 = 0$$ 6. **Simplify by dividing both sides by 3:** $$\cancel{3}y^2 + \cancel{3} \cdot 2 y - \cancel{3} \cdot 8 = 0 \implies y^2 + 2y - 8 = 0$$ 7. **Factor the quadratic:** $$(y + 4)(y - 2) = 0$$ 8. **Solve for $y$:** $$y + 4 = 0 \implies y = -4 \quad \text{(not valid since } y=2^x > 0)$$ $$y - 2 = 0 \implies y = 2$$ 9. **Back-substitute for $x$:** $$2^x = 2$$ 10. **Solve for $x$:** $$2^x = 2^1 \implies x = 1$$ **Final answer:** $$\boxed{1}$$