উদাহরণ ১৫৷
হলে দেখাও যে 
[H.S ‘92]
সমাধান:
মনে করি 
অতএব
 \to \left( 1 \right)-br /-\end{array}.jpg "\begin{array}{l}<br />\log x = k\left( {ry - qz} \right)\\<br /> \Rightarrow p\log x = kp\left( {ry - qz} \right)\\<br /> \Rightarrow \log {x^p} = k\left( {pry - pqz} \right) \to \left( 1 \right)<br />\end{array}")
অনুরূপে
 - k\left( {qrx - pry} \right) \to \left( 3 \right)-br /-\end{array}.jpg "\begin{array}{l}<br />\log {y^q} = kq\left( {pz - rx} \right) = k\left( {pqz - qrx} \right) \to \left( 2 \right)\\<br />\log {z^r} = kr\left( {qx - py} \right) = k\left( {qrx - pry} \right) \to \left( 3 \right)<br />\end{array}")
করে পাই
\log {x^p} + \log {y^q} + \log {z^r} = k\left( {rpy - qpz + pqz - qrx + qrx - rpy} \right)\\
\Rightarrow \log \left( {{x^p}{y^q}{z^r}} \right) = 0\\
\Rightarrow {x^p}{y^q}{z^r} = 1\left( {proved} \right)
\end{array}" alt="\begin{array}{l}
\log {x^p} + \log {y^q} + \log {z^r} = k\left( {rpy - qpz + pqz - qrx + qrx - rpy} \right)\\
\Rightarrow \log \left( {{x^p}{y^q}{z^r}} \right) = 0\\
\Rightarrow {x^p}{y^q}{z^r} = 1\left( {proved} \right)
\end{array}"/>