1. **State the problem:** Solve the exponential equation $$8^{x-1} = 6^{3x}$$. 2. **Recall the formula and rules:** To solve equations where the variable is in the exponent, we use logarithms. The key property is: $$\log(a^b) = b \log(a)$$ This allows us to bring the exponent down and solve for the variable. 3. **Apply logarithm to both sides:** $$\log\left(8^{x-1}\right) = \log\left(6^{3x}\right)$$ 4. **Use the logarithm power rule:** $$ (x-1) \log(8) = 3x \log(6) $$ 5. **Distribute and rearrange terms:** $$ x \log(8) - \log(8) = 3x \log(6) $$ 6. **Group terms with x on one side:** $$ x \log(8) - 3x \log(6) = \log(8) $$ 7. **Factor out x:** $$ x \left( \log(8) - 3 \log(6) \right) = \log(8) $$ 8. **Solve for x:** $$ x = \frac{\log(8)}{\log(8) - 3 \log(6)} $$ 9. **Simplify the expression:** Calculate the logarithms (using any base, commonly base 10 or natural log): - $$\log(8) = \log(2^3) = 3 \log(2)$$ - $$\log(6) = \log(2 \times 3) = \log(2) + \log(3)$$ Substitute back: $$ x = \frac{3 \log(2)}{3 \log(2) - 3 (\log(2) + \log(3))} = \frac{3 \log(2)}{3 \log(2) - 3 \log(2) - 3 \log(3)} = \frac{3 \log(2)}{-3 \log(3)} $$ 10. **Cancel common factors:** $$ x = \frac{\cancel{3} \log(2)}{-\cancel{3} \log(3)} = - \frac{\log(2)}{\log(3)} $$ 11. **Final answer:** $$ \boxed{x = - \frac{\log(2)}{\log(3)}} $$ This is the exact solution. You can approximate numerically if needed.