Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Chapter Test - Page 372: 17

Answer

$\frac{1\pm\sqrt {13}}{2}\approx- 1.303, 2.303$.

Work Step by Step

1. $log(x^2+3)=log(x+6) \Longrightarrow x^2+3=x+6 \Longrightarrow x^2-x-3=0 \Longrightarrow x=\frac{1\pm\sqrt {1-4(-3)}}{2}=\frac{1\pm\sqrt {13}}{2}\approx- 1.303, 2.303$. 2. Check, $x=\frac{1\pm\sqrt {13}}{2}$ fit(s).
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