## College Algebra (11th Edition)

$\log_3 \left(\dfrac{1}{32p^{5}}\right)$
$\bf{\text{Solution Outline:}}$ Use the Laws of Logarithms to write the given expression, $-\dfrac{3}{4}\log_3 16p^4-\dfrac{2}{3}\log_3 8p^3 ,$ 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_3 (16p^4)^{-3/4}+\log_3 (8p^3)^{-2/3} .\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_3 \dfrac{1}{(16p^4)^{3/4}}+\log_3 {\dfrac{1}{(8p^3)^{2/3}}} .\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_3 \dfrac{1}{\left(\sqrt[4]{16p^4}\right)^{3}}+\log_3 {\dfrac{1}{\left(\sqrt[3]{8p^3}\right)^{2}}} \\\\= \log_3 \dfrac{1}{\left(\sqrt[4]{(2p)^4}\right)^{3}}+\log_3 {\dfrac{1}{\left(\sqrt[3]{(2p)^3}\right)^{2}}} \\\\= \log_3 \dfrac{1}{\left(2p\right)^{3}}+\log_3 {\dfrac{1}{\left(2p\right)^{2}}} \\\\= \log_3 \dfrac{1}{8p^3}+\log_3 {\dfrac{1}{4p^2}} .\end{array} Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log_3 \left(\dfrac{1}{8p^3}\cdot\dfrac{1}{4p^2}\right) \\\\= \log_3 \dfrac{1}{32p^3p^2} .\end{array} Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \log_3 \dfrac{1}{32p^{3+2}} \\\\= \log_3 \left(\dfrac{1}{32p^{5}}\right) .\end{array}