Intermediate Algebra (6th Edition)

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

Chapter 9 - Review: 100

Answer

$x=\frac{1}{3}(\frac{log(9)}{log(4)}-2)\approx-.1383$

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. $4^{3x+2}=9$ Take the common logarithm of both sides (which has base 10). $log(4^{3x+2})=log(9)$ Use the power property of logarithms. $(3x+2) log(4)=log(9)$ Divide both sides by $log(4)$. $3x+2=\frac{log(9)}{log(4)}$ Subtract 2 from both sides. $3x=\frac{log(9)}{log(4)}-2$ Divide both sides by 3. $x=\frac{1}{3}(\frac{log(9)}{log(4)}-2)\approx-.1383$
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