## College Algebra (11th Edition)

$A=\dfrac{B}{x^C}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\log A=\log B-C\log x ,$ in terms of $A ,$ use the properties of logarithms to simplify and isolate the needed variable. $\bf{\text{Solution Details:}}$ Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log A=\log B-\log x^C .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log A=\log \dfrac{B}{x^C} .\end{array} Since both sides are expressed in a logarithm with the same base, then the logarithm may be dropped. Hence, the equation above becomes \begin{array}{l}\require{cancel} A=\dfrac{B}{x^C} .\end{array}