Answer
$A=\dfrac{B}{x^C}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}