Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.5 - Exponential and Logarithmic Equations - Exercise Set - Page 727: 90

Answer

The solution set is $\{3\}$

Work Step by Step

The given equation is $\log_2(x-1)-\log_2(x+3)=\log_2\left ( \frac{1}{x} \right )$ Use quotient rule. $\log_2 \left ( \frac{x-1}{x+3}\right )=\log_2\left ( \frac{1}{x} \right )$ $\frac{x-1}{x+3}= \frac{1}{x} $ Multiply both sides by $x(x+3)$. $x(x+3)\cdot \frac{x-1}{x+3}= x(x+3)\cdot \frac{1}{x} $ Simplify. $x(x-1)= (x+3)$ $x^2-x= x+3$ Subtract $x+3$ from both sides. $x^2-x-(x+3)= x+3-(x+3)$ $x^2-x-x-3= x+3-x-3$ $x^2-2x-3= 0$ Factor. $x^2+x-3x-3= 0$ $x(x+1)-3(x+1)= 0$ $(x+1)(x-3)= 0$ Set both factors equal to zero. $x+1=0$ or $x-3 =0$ Isolate $x$. $x=-1$ or $x=3$ Check for the solution. For $x=-1$. $\log_2(-1-1)-\log_2(-1+3)=\log_2\left ( \frac{1}{-1} \right )$ $\log_2(-2)-\log_2(2)=\log_2\left ( -1 \right )$ Not a solution. For $x=3$. $\log_2(3-1)-\log_2(3+3)=\log_2\left ( \frac{1}{3} \right )$ $\log_2(2)-\log_2(6)=\log_2\left ( \frac{1}{3} \right )$ $\log_2\left (\frac{2}{6}\right)=\log_2\left ( \frac{1}{3} \right )$ $\log_2\left (\frac{1}{3}\right)=\log_2\left ( \frac{1}{3} \right )$
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