Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Section 12.8 - Exponential and Logarithmic Equations and Problem Solving - Exercise Set: 29

Answer

$x=\dfrac{2}{3}$

Work Step by Step

$\log_{2}x+\log_{2}(3x+1)=1$ Combine $\log_{2}x+\log_{2}(3x+1)$ as the $\log$ of a product: $\log_{2}x(3x+1)=1$ $\log_{2}(3x^{2}+x)=1$ Rewrite in exponential form: $2^{1}=3x^{2}+x$ $3x^{2}+x=2$ Take the $2$ to the left side of the equation: $3x^{2}+x-2=0$ Solve this equation by factoring: $(x+1)(3x-2)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x+1=0$ $x=-1$ $3x-2=0$ $3x=2$ $x=\dfrac{2}{3}$ The original equation is undefined for $x=-1$, so the answer is just $x=\dfrac{2}{3}$
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