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: 17

Answer

$ln[\frac{\sqrt x}{(x+1)}]$

Work Step by Step

1. Use logarithmic properties $ln(pq) = lnp+lnq$, $ln(\frac{p}{q}) = lnp-lnq$ and $ln(p)^{m}= m lnp$. $\frac{1}{3}(x+2)^{3}+\frac{1}{2}[lnx-ln(x^{2}+3x+2)^{2}]=\frac{1}{3}.3ln(x+2)+\frac{1}{2}lnx-\frac{1}{2}.2ln(x^{2}+3x+2)$ $=ln[(x+2)\times\sqrt x]-ln(x+1)(x+2)$ 2. Use logarithmic property $ln(\frac{p}{q}) = lnp-lnq$ $\frac{1}{3}(x+2)^{3}+\frac{1}{2}[lnx-ln(x^{2}+3x+2)^{2}]$=$ln[\frac{(x+2)\sqrt x}{(x+1)(x+2)}]$ Hence, $\frac{1}{3}(x+2)^{3}+\frac{1}{2}[lnx-ln(x^{2}+3x+2)^{2}]=ln[\frac{\sqrt x}{(x+1)}]$
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