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: 23

Answer

$x=2$

Work Step by Step

$\log_{4}x+\log_{4}(x+6)=2$ Combine $\log_{4}x+\log_{4}(x+6)$ as the $\log$ of a product: $\log_{4}x(x+6)=2$ $\log_{4}(x^{2}+6x)=2$ Rewrite in exponential form: $4^{2}=x^{2}+6x$ $x^{2}+6x=16$ Take the $16$ to the left side of the equation: $x^{2}+6x-16=0$ Solve this equation by factoring: $(x+8)(x-2)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x+8=0$ $x=-8$ $x-2=0$ $x=2$ The initial equation is undefined for $x=-8$, so the answer to this is just $x=2$
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