Cho \({\log _8}\left| x \right| + {\log _4}{y^2} = 5\) và \({\log _8}\left| y \right| + {\log _4}{x^2} = 7\).

Điều kiện: \(x,y \ne 0\)

Theo đề bài ta có hệ phương trình:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{\log _8}\left| x \right| + {\log _4}{y^2} = 5\\
{\log _8}\left| y \right| + {\log _4}{x^2} = 7
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\frac{1}{3}{\log _2}\left| x \right| + {\log _2}\left| y \right| = 5\\
\frac{1}{3}{\log _2}\left| y \right| + {\log _2}\left| x \right| = 7
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{\log _2}\left| x \right| + {\log _2}{\left| y \right|^3} = 15\\
{\log _2}\left| y \right| + {\log _2}{\left| x \right|^3} = 21
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\log _2}\left| {x{y^3}} \right| = 15\\
{\log _2}\left| {{x^3}y} \right| = 21
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left| {x{y^3}} \right| = {2^{15}}{\rm{ (*)}}\\
\left| {{x^3}y} \right| = {2^{21}}
\end{array} \right. \Leftrightarrow \frac{{\left| {{x^3}y} \right|}}{{\left| {x{y^3}} \right|}} = 64 \Leftrightarrow {\left| {\frac{x}{y}} \right|^2} = 64 \Leftrightarrow \left| {\frac{x}{y}} \right| = 8 \Leftrightarrow \left| x \right| = 8\left| y \right|
\end{array}\)

Thay vào (*) ta có \(8{y^4} = {2^{15}} \Leftrightarrow \left| y \right| = \sqrt[4]{{4096}} = 8\)

Khi đó ta có \(P = \left| x \right| - \left| y \right| = 8\left| y \right| - \left| y \right| = 7\left| y \right| = 7.8 = 56\)