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: 61



Work Step by Step

Using the Quotient Property of logarithms, the given expression, $ \log_2(x^2-100)-\log_2(x+10)=1 $ is equivalent to \begin{array}{l}\require{cancel} \log_2\dfrac{x^2-100}{x+10}=1 .\end{array} Changing to exponential form, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{x^2-100}{x+10}=2^1 \\\\ \dfrac{x^2-100}{x+10}=2 .\end{array} Using the properties of equality and the concepts of quadratic equations, the solution/s to the equation above is/are \begin{array}{l}\require{cancel} x^2-100=2(x+10) \\\\ x^2-100=2x+20 \\\\ x^2-2x-100-20=0 \\\\ x^2-2x-120=0 \\\\ (x-12)(x+10)=0 .\end{array} Equating each factor to zero and then solving for the variable, the solutions are $ x=\left\{ -10,12 \right\} .$ Upon checking, only $ x=12 $ satisfies the original equation.
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