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 - Section 4.6 Logarithmic and Exponential Equations - 4.6 Assess Your Understanding - Page 337: 30



Work Step by Step

Apply the logarithmic property: $\log_a(\dfrac{M}{N}) = \log_a M-\log_a N$ and rearrange the given expression to obtain: $\log_{4}[\dfrac{x^2 -9}{x+3}] =3$ Since $\log_m{n} = 1 $ gives: $m^{(1)}=n$, then we have: $4^3=\dfrac{x^2-9}{x+3}$ $x^2-9=64 x+192$ or, $x^2-64x -201=0$ This is a quadratic equation; thus by factoring, it will become: $(x+3)(x-67)=0$ By the zero product property, we have: $x=-3$ and $x=67$ Since the domain of the variable is $x \gt 0$, we cannot accept the value of $x=-3$ Thus, our answer is: $x=67$
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