Intermediate Algebra (6th Edition)

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

Chapter 9 - Section 9.7 - Common Logarithms, Natural Logarithms, and Change of Base - Exercise Set: 45



Work Step by Step

We are given the equation $\log(2x+1)=-.5$. To solve for x, remember that the base of a common logarithm is understood to be 10. Therefore, $log(2x+1)=log_{10}2x+1=-.5$. 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, $2x+1=10^{-.5}$. Subtract 1 from both sides. $2x=10^{-.5}-1$ Divide both sides by 2. $x=\frac{10^{-.5}-1}{2}\approx-.3419$
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