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: 46



Work Step by Step

We are given the equation $\log(3x-2)=-.8$. To solve for x, remember that the base of a common logarithm is understood to be 10. Therefore, $log(3x-2)=log_{10}(3x-2)=-.8$. 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, $3x-2=10^{-.8}$. Add 2 to both sides. $3x=10^{-.8}+2$ Divide both sides by 3. $x=\frac{10^{-.8}+2}{3}\approx.7195$
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