College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.3 - Logarithmic Functions - 4.3 Exercises: 80

Answer

$\log_y \dfrac{\sqrt[6]{p^5}}{p^{2}}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Laws of Logarithms to write the given expression, $ \dfrac{1}{2}\log_y p^3q^4-\dfrac{2}{3}\log_y p^4q^3 ,$ as a single logarithm. $\bf{\text{Solution Details:}}$ Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log_y (p^3q^4)^{1/2}-\log_y (p^4q^3)^{2/3} .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log_y \dfrac{(p^3q^4)^{1/2}}{(p^4q^3)^{2/3}} .\end{array} Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_y \dfrac{p^{3\cdot\frac{1}{2}}q^{4\cdot\frac{1}{2}}}{p^{4\cdot\frac{2}{3}}q^{3\cdot\frac{2}{3}}} \\\\= \log_y \dfrac{p^{\frac{3}{2}}q^{2}}{p^{\frac{8}{3}}q^{2}} \\\\= \log_y \dfrac{p^{\frac{3}{2}}\cancel{q^{2}}}{p^{\frac{8}{3}}\cancel{q^{2}}} \\\\= \log_y \dfrac{p^{\frac{3}{2}}}{p^{\frac{8}{3}}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} \log_y p^{\frac{3}{2}-\frac{8}{3}} \\\\= \log_y p^{\frac{9}{6}-\frac{16}{6}} \\\\= \log_y p^{-\frac{7}{6}} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_y \dfrac{1}{p^{\frac{7}{6}}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_y \dfrac{1}{\sqrt[6]{p^7}} .\end{array} Rationalizing the denominator results to \begin{array}{l}\require{cancel} \log_y \dfrac{1}{\sqrt[6]{p^7}}\cdot\dfrac{\sqrt[6]{p^5}}{\sqrt[6]{p^5}} \\\\= \log_y \dfrac{\sqrt[6]{p^5}}{\sqrt[6]{p^{12}}} \\\\= \log_y \dfrac{\sqrt[6]{p^5}}{\sqrt[6]{(p^{2})^6}} \\\\= \log_y \dfrac{\sqrt[6]{p^5}}{p^{2}} .\end{array}
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