1. **Stating the problem:** Solve the equation $$25^x \cdot \left(\frac{1}{5}\right)^{x-1} = \left(\frac{1}{5}\right)^{-x}$$ for $x$. 2. **Rewrite the bases:** Note that $25 = 5^2$, so rewrite the equation using base 5: $$\left(5^2\right)^x \cdot \left(\frac{1}{5}\right)^{x-1} = \left(\frac{1}{5}\right)^{-x}$$ 3. **Apply power of a power rule:** $$5^{2x} \cdot 5^{-(x-1)} = 5^{x}$$ Here, we used $\left(\frac{1}{5}\right)^a = 5^{-a}$. 4. **Combine the left side exponents:** $$5^{2x} \cdot 5^{-x+1} = 5^{2x + (-x + 1)} = 5^{x+1}$$ 5. **Set the equation:** $$5^{x+1} = 5^{x}$$ 6. **Since bases are equal and positive (not 1), set exponents equal:** $$x + 1 = x$$ 7. **Solve for $x$:** $$x + 1 = x \implies 1 = 0$$ This is a contradiction, meaning no solution from this step. 8. **Re-examine step 3:** The right side was rewritten as $5^{x}$, but original right side is $\left(\frac{1}{5}\right)^{-x} = 5^{x}$, which is correct. 9. **Check the original equation carefully:** $$25^x \cdot \left(\frac{1}{5}\right)^{x-1} = \left(\frac{1}{5}\right)^{-x}$$ Rewrite as: $$5^{2x} \cdot 5^{-(x-1)} = 5^{x}$$ Simplify left exponent: $$5^{2x - x + 1} = 5^{x}$$ $$5^{x + 1} = 5^{x}$$ 10. **Set exponents equal:** $$x + 1 = x$$ 11. **This is impossible, so no solution exists.** **Final answer:** There is no value of $x$ that satisfies the equation.