1. **State the problem:** Solve the equation $$\frac{8^x - 2^x}{6^x - 3^x} = 2$$ for $x$. 2. **Rewrite the bases in terms of prime factors:** - $8 = 2^3$ - $6 = 2 \times 3$ So, $$8^x = (2^3)^x = 2^{3x}$$ $$6^x = (2 \times 3)^x = 2^x \cdot 3^x$$ 3. **Substitute these into the equation:** $$\frac{2^{3x} - 2^x}{2^x \cdot 3^x - 3^x} = 2$$ 4. **Factor terms where possible:** - Numerator: $2^x(2^{2x} - 1)$ - Denominator: $3^x(2^x - 1)$ So the equation becomes: $$\frac{2^x(2^{2x} - 1)}{3^x(2^x - 1)} = 2$$ 5. **Rewrite the equation:** $$\frac{2^x(2^{2x} - 1)}{3^x(2^x - 1)} = 2$$ Multiply both sides by denominator: $$2^x(2^{2x} - 1) = 2 \cdot 3^x (2^x - 1)$$ 6. **Let $a = 2^x$ and $b = 3^x$ for simplicity:** $$a(a^2 - 1) = 2b(a - 1)$$ 7. **Expand both sides:** $$a^3 - a = 2b(a - 1)$$ 8. **Rewrite right side:** $$a^3 - a = 2ab - 2b$$ 9. **Bring all terms to one side:** $$a^3 - a - 2ab + 2b = 0$$ 10. **Group terms:** $$a^3 - a - 2ab + 2b = (a^3 - a) - 2b(a - 1) = 0$$ 11. **Isolate $b$:** $$2b(a - 1) = a^3 - a$$ $$b = \frac{a^3 - a}{2(a - 1)}$$ 12. **Recall $b = 3^x$ and $a = 2^x$, so:** $$3^x = \frac{(2^x)^3 - 2^x}{2(2^x - 1)} = \frac{2^{3x} - 2^x}{2(2^x - 1)}$$ 13. **Try to find $x$ by testing values:** - For $x=1$: $$LHS = \frac{8^1 - 2^1}{6^1 - 3^1} = \frac{8 - 2}{6 - 3} = \frac{6}{3} = 2$$ This satisfies the equation. 14. **Check if other solutions exist:** Because the equation is transcendental, and $x=1$ satisfies it, this is the solution. **Final answer:** $$x = 1$$