College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Test - Page 472: 18


$x=\left\{ 0,6 \right\}$

Work Step by Step

Changing to exponential form, the given equation, $ \log_2(x-4)(x-2)=3 ,$ is equivalent to \begin{array}{l}\require{cancel} (x-4)(x-2)=2^3 \\\\ (x-4)(x-2)=8 .\end{array} Using the concepts of solving quadratic equations, then \begin{array}{l}\require{cancel} x(x)+x(-2)-4(x)-4(-2)=8 \\\\ x^2-2x-4x+8=8 \\\\ x^2-2x-4x=8-8 \\\\ x^2-6x=0 \\\\ x(x-6)=0 .\end{array} Equating each factor to $0$ and solving for the variable, the solutions are $ x=\left\{ 0,6 \right\} .$ Upon checking, $ x=\left\{ 0,6 \right\} $ satisfy the original equation.
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