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: 50

Answer

$x=\left\{ -\dfrac{8}{3},0 \right\}$

Work Step by Step

Changing to exponential form, the given expression, $ \log_5[(3x+5)(x+1)]=1 ,$ is equivalent to \begin{array}{l}\require{cancel} (3x+5)(x+1)=5^1 .\end{array} Using concepts of quadratic equations, the solution/s to the equation above is/are \begin{array}{l}\require{cancel} 3x(x)+3x(1)+5(x)+5(1)=5 \\\\ 3x^2+3x+5x+5-5=0 \\\\ 3x^2+8x=0 \\\\ x(3x+8)=0 .\end{array} Equating each factor to zero and then solving for the variable, the solutions are $ x=\left\{ -\dfrac{8}{3},0 \right\} .$ Upon checking, $ x=\left\{ -\dfrac{8}{3},0 \right\} $ satisfy the original equation.
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