Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - Summary Exercises on Functions: Domains and Defining Equations - Exercises - Page 487: 23

Answer

$(-\infty,-2)\cup(-2,3)\cup(3,\infty)$.

Work Step by Step

The domain requirement for $f(x)=log(\frac{x+2}{x-3})^2$ is $x-3\ne0, x+2\ne0$ which gives $x\ne-2,3$ or in interval notation $(-\infty,-2)\cup(-2,3)\cup(3,\infty)$.
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