Intermediate Algebra (6th Edition)

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

Chapter 9 - Review: 90



Work Step by Step

We are given the equation $\ln(3x+1)=2$. To solve for x, remember that the base of a natural logarithm is understood to be $e$. Therefore, $\ln(3x+1)=log_{e}(3x+1)=2$. If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x\gt0$ and every real number $y$. Therefore, $3x+1=e^{2}$. Subtract 1 from both sides. $3x=e^{2}-1$ Divide both sides by 3. $x=\frac{e^{2}-1}{3}$
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