## Calculus 8th Edition

$ln3+\frac{1}{3}ln8=ln6$
Use logarithmic properties $ln(pq) = lnp+lnq$, $ln(\frac{p}{q}) = lnp-lnq$ and $ln(p)^{m}= m lnp$. $ln3+\frac{1}{3}ln8=ln3+ln(8)^{\frac{1}{3}}=ln(3\times\sqrt[3] 8)$ Use logarithmic property $ln(pq) = lnp+lnq$, we get $ln(3\times2)=ln6$ Hence, $ln3+\frac{1}{3}ln8=ln6$