Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.3 Logarithmic Functions - 6.3 Exercises: 18

Answer

$lnb+2lnc-3lnd=ln[\frac{bc^{2}}{d^{3}}]$

Work Step by Step

To express the quantity as a single quantity we need to follow some steps. 1.Use logarithmic properties $ln(pq) = lnp+lnq$ and $ln(p)^{m}= m lnp$. $lnb+2lnc-3lnd=lnb+lnc^{2}-lnd^{3}$ $= ln(bc^{2})-ln(d^{3})$ $=ln[\frac{bc^{2}}{d^{3}}]$ Hence, $lnb+2lnc-3lnd=ln[\frac{bc^{2}}{d^{3}}]$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.