উদাহরণ ১৬৷[tex]x = {\log _a}\left( {bc} \right),y = {\log _b}\left( {ca} \right),z = {\log _c}\left( {ab} \right)[/tex] হলে দেখাও যে
[tex]\begin{array}{l}
\frac{1}{{x + 1}} + \frac{1}{{y + 1}} + \frac{1}{{z + 1}} = 1\\
x + y + z + 2 = xyz
\end{array}[/tex] [H.S ’91,’93,’94]
প্রমান:
[tex]\begin{array}{l}
x + 1\\
= {\log _a}\left( {bc} \right) + 1\\
= {\log _a}\left( {bc} \right) + {\log _a}a\\
= {\log _a}\left( {abc} \right)\\
\Rightarrow \frac{1}{{x + 1}} = \frac{1}{{{{\log }_a}\left( {abc} \right)}} = {\log _{abc}}a \to \left( 1 \right)
\end{array}[/tex]
অনুরূপে
[tex]\begin{array}{l}
\frac{1}{{y + 1}} = {\log _{abc}}b \to \left( 2 \right)\\
\frac{1}{{z + 1}} = {\log _{abc}}c \to \left( 3 \right)
\end{array}[/tex]
(1)+ (2) + (3) করে পাই
[tex]\begin{array}{l}
\frac{1}{{x + 1}} + \frac{1}{{y + 1}} + \frac{1}{{z + 1}}\\
= {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c\\
= {\log _{abc}}\left( {abc} \right)\\
= 1\left( {proved} \right)
\end{array}[/tex]
[tex]\begin{array}{l}
\frac{1}{{x + 1}} + \frac{1}{{y + 1}} + \frac{1}{{z + 1}} = 1\\
\Rightarrow \frac{{x + 1 + y + 1}}{{\left( {x + 1} \right)\left( {y + 1} \right)}} = \frac{z}{{\left( {z + 1} \right)}}\\
\Rightarrow x + y + 2 = \frac{{z\left( {xy + x + y + 1} \right)}}{{z + 1}}\\
\Rightarrow xz + yz + 2z + x + y + 2 = xyz + xz + yz + z\\
\Rightarrow x + y + z + 2 = xyz\left( {proved} \right)
\end{array}[/tex]
