Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 9 - Review - Page 598: 99



Work Step by Step

According to the logarithm property of equality $log_{b}a=log_{b}c$ is equivalent to $a=c$ (where a, b, and c are real numbers such that $log_{b}a$ and $log_{b}c$ are real numbers and $b\ne1$). We can use this property to solve for x. $3^{2x+1}=6$ Take the common logarithm of both sides (which has base 10). $log(3^{2x+1})=log(6)$ Use the power property of logarithms. $(2x+1) log(3)=log(6)$ Divide both sides by $log(3)$. $2x+1=\frac{log(6)}{log(3)}$ Subtract 1 from both sides. $2x=\frac{log(6)}{log(3)}-1$ Divide both sides by 2. $x=\frac{1}{2}(\frac{log(6)}{log(3)-1})\approx.3155$
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