College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 447: 47


$x=\{ -6,4 \}$

Work Step by Step

Changing to exponential form, the given expression, $ \log_3[(x+5)(x-3)]=2 ,$ is equivalent to \begin{array}{l}\require{cancel} (x+5)(x-3)=3^2 .\end{array} Using concepts of quadratic equations, the solution/s to the equation above is/are \begin{array}{l}\require{cancel} x(x)+x(-3)+5(x)+5(-3)=9 \\\\ x^2-3x+5x-15-9=0 \\\\ x^2+2x-24=0 \\\\ (x+6)(x-4)=0 .\end{array} Equating each factor to zero and then solving for the variable, the solutions are $ x=\{ -6,4 \} .$
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