1. **State the problem:** Solve the equation $$\left(\frac{1}{125}\right)^x - \left(25^{\frac{3}{4}}\right)^{-1} = 0$$ for $x$. 2. **Rewrite the equation:** We want to find $x$ such that $$\left(\frac{1}{125}\right)^x = \left(25^{\frac{3}{4}}\right)^{-1}$$ 3. **Express bases as powers of prime numbers:** - Note that $125 = 5^3$, so $$\frac{1}{125} = 5^{-3}$$ - Also, $25 = 5^2$, so $$25^{\frac{3}{4}} = \left(5^2\right)^{\frac{3}{4}} = 5^{2 \times \frac{3}{4}} = 5^{\frac{3}{2}}$$ 4. **Simplify the right side:** $$\left(25^{\frac{3}{4}}\right)^{-1} = \left(5^{\frac{3}{2}}\right)^{-1} = 5^{-\frac{3}{2}}$$ 5. **Rewrite the equation with the same base:** $$\left(5^{-3}\right)^x = 5^{-\frac{3}{2}}$$ 6. **Use the power of a power rule:** $$5^{-3x} = 5^{-\frac{3}{2}}$$ 7. **Since the bases are equal and nonzero, set exponents equal:** $$-3x = -\frac{3}{2}$$ 8. **Solve for $x$:** $$x = \frac{-\frac{3}{2}}{-3} = \frac{3/2}{3} = \frac{3}{2} \times \frac{1}{3} = \frac{1}{2}$$ **Final answer:** $$x = \frac{1}{2}$$