## Precalculus (6th Edition) Blitzer

$$4x^3$$
By applying the quotient rule and the identity $b^{\log_bM}=M$, we have$$e^{\ln 8x^5- \ln 2x^2}=e^{\ln \left ( \frac{8x^5}{2x^2} \right )}=\frac{8x^5}{2x^2}=4x^3.$$