1. **State the problem:** We need to find the value of $x$ in the equation $$5 \times 625^{3-x} = \left(\frac{1}{25}\right)^{3x}.$$\n\n2. **Rewrite the bases as powers of 5:** Note that $625 = 5^4$ and $25 = 5^2$. So the equation becomes $$5 \times (5^4)^{3-x} = \left(5^{-2}\right)^{3x}.$$\n\n3. **Simplify the exponents:** Using the power of a power rule $(a^m)^n = a^{mn}$, we get $$5 \times 5^{4(3-x)} = 5^{-2(3x)}.$$\nThis simplifies to $$5 \times 5^{12 - 4x} = 5^{-6x}.$$\n\n4. **Combine the left side:** Since $5 = 5^1$, multiply powers with the same base by adding exponents: $$5^{1 + 12 - 4x} = 5^{-6x}$$ which is $$5^{13 - 4x} = 5^{-6x}.$$\n\n5. **Set the exponents equal:** Because the bases are the same and nonzero, the exponents must be equal: $$13 - 4x = -6x.$$\n\n6. **Solve for $x$:** Add $6x$ to both sides: $$13 + 2x = 0.$$\nSubtract 13 from both sides: $$2x = -13.$$\nDivide both sides by 2: $$x = -\frac{13}{2} = -6.5.$$\n\n**Final answer:** $$x = -6.5.$$