Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 368: 65

Answer

$x=6$

Work Step by Step

$\log_{9}(x-5)+\log_{9}(x+3)=1$ Combine the logarithms on the left side of the equation as a product: $\log_{9}(x-5)(x+3)=1$ $\log_{9}(x^{2}-2x-15)=1$ Write this equation in exponential form: $x^{2}-2x-15=9^{1}$ $x^{2}-2x-15=9$ Take the $9$ to the left side of the equation: $x^{2}-2x-15-9=0$ $x^{2}-2x-24=0$ Solve by factoring: $(x+4)(x-6)=0$ We get two solutions: $x=-4$ and $x=6$ We see that the initial equation is undefined when $x=-4$, so we can discard that solution. Our final answer is $x=6$.
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