1. **State the problem:** We need to find the value of $n$ such that $$Pint_2^{n-3} = 20 imes Pint_3^n.$$ 2. **Understand the notation:** Assuming $Pint_a^b$ means $a^b$, the equation becomes $$2^{n-3} = 20 imes 3^n.$$ 3. **Rewrite the equation:** $$2^{n-3} = 20 imes 3^n$$ 4. **Express $2^{n-3}$ as $\frac{2^n}{2^3}$:** $$\frac{2^n}{8} = 20 imes 3^n$$ 5. **Multiply both sides by 8:** $$2^n = 160 imes 3^n$$ 6. **Divide both sides by $3^n$:** $$\frac{2^n}{3^n} = 160$$ 7. **Rewrite the left side as $(\frac{2}{3})^n$:** $$(\frac{2}{3})^n = 160$$ 8. **Take the natural logarithm of both sides:** $$n \ln(\frac{2}{3}) = \ln(160)$$ 9. **Solve for $n$:** $$n = \frac{\ln(160)}{\ln(\frac{2}{3})}$$ 10. **Calculate the values:** $$\ln(160) \approx 5.0752$$ $$\ln(\frac{2}{3}) = \ln(2) - \ln(3) \approx 0.6931 - 1.0986 = -0.4055$$ 11. **Final value:** $$n = \frac{5.0752}{-0.4055} \approx -12.52$$ **Answer:** $n \approx -12.52$