Start by creating the system of constraint equations.

\begin{equation*}
\left\{ \begin{matrix} A \cdot 2^p = 5 \\ A \cdot 5^p = 1 \end{matrix} \right.
\end{equation*}

To use the method of substitution, it is easiest to solve for \(A\text{:}\)

\begin{gather*}
A \cdot 2^p = 5, \\
A = \frac{5}{2^p}.
\end{gather*}

Substitute this into the second equation to get an equation for \(p\text{:}\)

\begin{gather*}
A \cdot 5^p = 1 \\
\Big( \frac{5}{\displaystyle 2^p} \Big) \cdot 5^p = 1 \\
5 \cdot \Big( \frac{5^p}{2^p} \Big) = 1 \\
5 \cdot (\textstyle \frac{5}{2})^p = 1
\end{gather*}

To solve this, we isolate the *exponential* with base \(\frac{5}{2}\) and then use a logarithm as the inverse operation.

\begin{gather*}
5 \cdot (\textstyle \frac{5}{2})^p = 1 \\
(\textstyle \frac{5}{2})^p = \frac{1}{5} \\
\textstyle \log_{\frac{5}{2}}(\exp_{\frac{5}{2}}(p)) = \log_{\frac{5}{2}}(\frac{1}{5})\\
\textstyle p = \log_{\frac{5}{2}}(\frac{1}{5})\\
p = \frac{\log(\frac{1}{5})}{\log(\frac{5}{2})} \approx -1.7565
\end{gather*}

This allows us to find the value of \(A\text{:}\)

\begin{equation*}
A = \frac{5}{2^p} \approx \frac{5}{2^{-1.7565}} \approx 16.894.
\end{equation*}

The exact function is defined by

\begin{equation*}
f(x) = \frac{5}{2^p} \cdot x^p, \qquad p = \log_{\frac{5}{2}}({\textstyle \frac{1}{5}}),
\end{equation*}

with an approximate form

\begin{equation*}
f(x) \approx 16.894 \cdot x^{-1.7565}.
\end{equation*}