Intermediate Algebra (6th Edition)

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

Chapter 9 - Section 9.6 - Properties of Logarithms - Exercise Set: 45

Answer

$\frac{1}{2}log_{b}7+\frac{1}{2}log_{b}x$

Work Step by Step

The product property of logarithms tells us that $log_{b}xy=log_{b}x+log_{b}y$ (where x, y, and, b are positive real numbers and $b\ne1$). Therefore, $log_{b}\sqrt(7x)= log_{b}7^{\frac{1}{2}}x^{\frac{1}{2}}= log_{b}7^{\frac{1}{2}}+log_{b}x^{\frac{1}{2}}$. The power property of logarithms tells us that $log_{b}x^{r}=r log_{b}x$ (where x and b are positive real numbers, $b\ne1$, and r is a real number). Therefore, $ log_{b}7^{\frac{1}{2}}+log_{b}x^{\frac{1}{2}} =\frac{1}{2}log_{b}7+\frac{1}{2}log_{b}x$.
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