## Calculus 8th Edition

Published by Cengage

# Chapter 6 - Inverse Functions - 6.3 Logarithmic Functions - 6.3 Exercises - Page 426: 16

#### Answer

$ln3+\frac{1}{3}ln8=ln6$

#### Work Step by Step

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, $ln3+\frac{1}{3}ln8=ln6$

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