উদাহরণ ১৭৷ [tex]x = {\log _{2a}}a,y = {\log _{3a}}2a,z = {\log _{4a}}3a[/tex] হলে দেখাও যে [tex]xyz + 1 = 2yz[/tex] [H.S ‘97]
প্রমান:
[tex]\begin{array}{l}
xyz + 1\\
= {\log _{2a}}a \times {\log _{3a}}2a \times {\log _{4a}}3a + 1\\
= \left( {{{\log }_{2a}}a \times {{\log }_{3a}}2a} \right) \times {\log _{4a}}3a + 1\\
= {\log _{3a}}a \times {\log _{4a}}3a + 1\\
= {\log _{4a}}a + 1\\
= {\log _{4a}}a + {\log _{4a}}4a\\
= {\log _{4a}}\left( {4a \cdot a} \right)\\
= {\log _{4a}}4{a^2}\\
= 2{\log _{4a}}2a\\
= 2{\log _{4a}}3a \times {\log _{3a}}2a\\
= 2yz\\
\Rightarrow xyz + 1 = 2yz\left( {proved} \right)
\end{array}[/tex]
