1. **State the problem:** Solve algebraically for $x$ in the equation $$6^{3x+1} = 2^{2x-3}.$$\n\n2. **Recall the formula and rules:** To solve equations with different bases, take the logarithm of both sides. We can use the natural logarithm $\ln$ or common logarithm $\log$. Here, we use $\ln$.\n\n3. **Apply logarithm to both sides:**\n$$\ln\left(6^{3x+1}\right) = \ln\left(2^{2x-3}\right)$$\n\n4. **Use the power rule of logarithms:**\n$$ (3x+1)\ln(6) = (2x-3)\ln(2) $$\n\n5. **Expand both sides:**\n$$ 3x\ln(6) + \ln(6) = 2x\ln(2) - 3\ln(2) $$\n\n6. **Group terms with $x$ on one side and constants on the other:**\n$$ 3x\ln(6) - 2x\ln(2) = -3\ln(2) - \ln(6) $$\n\n7. **Factor out $x$:**\n$$ x\left(3\ln(6) - 2\ln(2)\right) = -3\ln(2) - \ln(6) $$\n\n8. **Solve for $x$:**\n$$ x = \frac{-3\ln(2) - \ln(6)}{3\ln(6) - 2\ln(2)} $$\n\n9. **Simplify the numerator and denominator if desired:**\nNumerator: $$ -3\ln(2) - \ln(6) = -3\ln(2) - \left(\ln(2) + \ln(3)\right) = -4\ln(2) - \ln(3) $$\nDenominator: $$ 3\ln(6) - 2\ln(2) = 3\left(\ln(2) + \ln(3)\right) - 2\ln(2) = 3\ln(2) + 3\ln(3) - 2\ln(2) = \ln(2) + 3\ln(3) $$\n\nSo,\n$$ x = \frac{-4\ln(2) - \ln(3)}{\ln(2) + 3\ln(3)} $$\n\n10. **Calculate decimal approximation:**\nUsing $\ln(2) \approx 0.6931$ and $\ln(3) \approx 1.0986$,\nNumerator: $$ -4(0.6931) - 1.0986 = -2.7724 - 1.0986 = -3.8710 $$\nDenominator: $$ 0.6931 + 3(1.0986) = 0.6931 + 3.2958 = 3.9889 $$\n\nTherefore,\n$$ x \approx \frac{-3.8710}{3.9889} \approx -0.97 $$\n\n**Final answer:**\n$$ x = \frac{-4\ln(2) - \ln(3)}{\ln(2) + 3\ln(3)} \approx -0.97 $$