College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 7 - Summary, Review, and Test - Cumulative Review Exercises (Chapters 1-7): 7

Answer

$x = 3$

Work Step by Step

$$log_{2}(x + 1) + log_{2}(x - 1) = 3$$ By using the Rules of Logarithmics, we can simplify the first two terms as follows: $$log_{2}(x + 1) + log_{2}(x - 1)$$ $$log_{2}(x+1)(x-1)$$ $$log_{2}(x^{2} - 1)$$ And use it to re-write the original equation as: $$log_{2}(x^{2} - 1) = 3$$ Finally, we can write the equation in exponential form and solve accordingly: $$2^{3} = (x^{2} - 1)$$ $8 = x^{2} - 1$ $8 + 1 = x^{2}$ $\frac{+}{} \sqrt {9} = \frac{+}{} 3 = x$ Since using $x= -3$ in the original equation would give us $\log_{2}(-2) + \log_{2}(-4) = 3$, an undefined logrithmic equation, the only possible solution is $x=3$.
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