1. **Problem:** Solve the equation $$\sqrt[4]{0.04^x} = 125^{x-2}$$. 2. **Formula and rules:** - Recall that $$\sqrt[4]{a} = a^{\frac{1}{4}}$$. - Express numbers as powers of primes for easier comparison. 3. **Rewrite the equation:** $$\sqrt[4]{0.04^x} = (0.04^x)^{\frac{1}{4}} = 0.04^{\frac{x}{4}}$$ 4. **Express 0.04 and 125 as powers:** - $$0.04 = \frac{4}{100} = \frac{2^2}{10^2} = \frac{2^2}{(2 \cdot 5)^2} = \frac{2^2}{2^2 \cdot 5^2} = 5^{-2}$$ - So, $$0.04 = 5^{-2}$$ - $$125 = 5^3$$ 5. **Rewrite the equation with powers of 5:** $$0.04^{\frac{x}{4}} = 125^{x-2} \implies (5^{-2})^{\frac{x}{4}} = (5^3)^{x-2}$$ 6. **Simplify exponents:** $$5^{-\frac{2x}{4}} = 5^{3(x-2)} \implies 5^{-\frac{x}{2}} = 5^{3x - 6}$$ 7. **Set exponents equal:** $$-\frac{x}{2} = 3x - 6$$ 8. **Solve for $$x$$:** Multiply both sides by 2: $$-x = 6x - 12$$ Add $$x$$ to both sides: $$0 = 7x - 12$$ Add 12 to both sides: $$12 = 7x$$ Divide both sides by 7: $$x = \frac{12}{7}$$ **Final answer:** $$x = \frac{12}{7}$$ (Option D)