Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 895: 70


The solution is, $3{{\log }_{b}}x-2{{\log }_{b}}5-\frac{1}{3}{{\log }_{b}}y={{\log }_{b}}\frac{{{x}^{3}}{{y}^{\frac{-1}{3}}}}{25}$.

Work Step by Step

First use the power rule for logarithms to make all the coefficients 1. That is, $a{{\log }_{b}}x={{\log }_{b}}{{x}^{a}}$ Thus, ${{\log }_{b}}{{x}^{3}}-{{\log }_{b}}{{5}^{2}}-{{\log }_{b}}{{y}^{\frac{1}{3}}}$ Simplifying it further, we get, ${{\log }_{b}}{{x}^{3}}-{{\log }_{b}}25-{{\log }_{b}}{{y}^{\frac{1}{3}}}$ ${{\log }_{b}}{{x}^{3}}-\left( {{\log }_{b}}25+{{\log }_{b}}{{y}^{\frac{1}{3}}} \right)$ Use the product rule for logarithms, ${{\log }_{b}}M+{{\log }_{b}}N={{\log }_{b}}\left( M\cdot N \right)$ Thus, ${{\log }_{b}}{{x}^{3}}-\left( {{\log }_{b}}25{{y}^{\frac{1}{3}}} \right)$ Now, we will use the quotient rule for logarithms, ${{\log }_{b}}\left( \frac{M}{N} \right)={{\log }_{b}}M-{{\log }_{b}}N$ Thus, ${{\log }_{b}}\frac{{{x}^{3}}}{25{{y}^{\frac{1}{3}}}}={{\log }_{b}}\frac{{{x}^{3}}{{y}^{\frac{-1}{3}}}}{25}$ Hence, $3{{\log }_{b}}x-2{{\log }_{b}}5-\frac{1}{3}{{\log }_{b}}y={{\log }_{b}}\frac{{{x}^{3}}{{y}^{\frac{-1}{3}}}}{25}$ Hence, $3{{\log }_{b}}x-2{{\log }_{b}}5-\frac{1}{3}{{\log }_{b}}y={{\log }_{b}}\frac{{{x}^{3}}{{y}^{\frac{-1}{3}}}}{25}$.
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