1. **State the problem:** Solve the equation $$6^{x-1} = 2^{3x+1}$$ and express the solution in the form $$\frac{\ln a}{\ln b}$$. 2. **Recall the formula and rules:** To solve equations with exponents, take the natural logarithm (ln) of both sides to bring down the exponents. 3. **Apply ln to both sides:** $$\ln\left(6^{x-1}\right) = \ln\left(2^{3x+1}\right)$$ 4. **Use the power rule of logarithms:** $$ (x-1)\ln 6 = (3x+1)\ln 2 $$ 5. **Expand both sides:** $$ x\ln 6 - \ln 6 = 3x\ln 2 + \ln 2 $$ 6. **Group terms with x on one side and constants on the other:** $$ x\ln 6 - 3x\ln 2 = \ln 6 + \ln 2 $$ 7. **Factor out x:** $$ x(\ln 6 - 3\ln 2) = \ln 6 + \ln 2 $$ 8. **Simplify the logarithmic expressions:** Recall that $$3\ln 2 = \ln 2^3 = \ln 8$$, so $$ \ln 6 - 3\ln 2 = \ln 6 - \ln 8 = \ln \frac{6}{8} = \ln \frac{3}{4} $$ Also, $$ \ln 6 + \ln 2 = \ln (6 \times 2) = \ln 12 $$ So the equation becomes: $$ x \ln \frac{3}{4} = \ln 12 $$ 9. **Solve for x:** $$ x = \frac{\ln 12}{\ln \frac{3}{4}} $$ 10. **Final answer:** $$ x = \frac{\ln 12}{\ln \frac{3}{4}} $$ This is the solution expressed in the form $$\frac{\ln a}{\ln b}$$ with $$a=12$$ and $$b=\frac{3}{4}$$.