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: 49


$x=\left\{ -8,0 \right\}$

Work Step by Step

Changing to exponential form, the given expression, $ \log_2[(2x+8)(x+4)]=5 ,$ is equivalent to \begin{array}{l}\require{cancel} (2x+8)(x+4)=2^5 .\end{array} Using concepts of quadratic equations, the solution/s to the equation above is/are \begin{array}{l}\require{cancel} 2x(x)+2x(4)+8(x)+8(4)=32 \\\\ 2x^2+8x+8x+32-32=0 \\\\ 2x^2+16x=0 \\\\ \dfrac{2x^2+16x}{2}=\dfrac{0}{2} \\\\ x^2+8x=0 \\\\ x(x+8)=0 .\end{array} Equating each factor to zero and then solving for the variable, the solutions are $ x=\left\{ -8,0 \right\} .$ Upon checking, $ x=\left\{ -8,0 \right\} $ satisfy the original equation.
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