উদাহরণ ৭৷ প্রমান করো [tex]\frac{1}{{{{\log }_a}bc + 1}} + \frac{1}{{{{\log }_b}ca + 1}} + \frac{1}{{{{\log }_c}ab + 1}} = 1[/tex] [H.S’99]
প্রমান:
[tex]\begin{array}{l}
\frac{1}{{{{\log }_a}bc + 1}}\\
= \frac{1}{{{{\log }_a}bc + {{\log }_a}a}}\\
= \frac{1}{{{{\log }_a}abc}}\\
= {\log _{abc}}a\\
\Rightarrow \frac{1}{{{{\log }_a}bc + 1}} = {\log _{abc}}a \to \left( 1 \right)
\end{array}[/tex]
অনুরূপে [tex]\frac{1}{{{{\log }_b}ca + 1}} = {\log _{abc}}b \to \left( 2 \right),\frac{1}{{{{\log }_c}ab + 1}} = {\log _{abc}}c \to \left( 3 \right)[/tex]
(1), (2), (3) থেকে পাই
[tex]\begin{array}{l}
\frac{1}{{{{\log }_a}bc + 1}} + \frac{1}{{{{\log }_b}ca + 1}} + \frac{1}{{{{\log }_c}ab + 1}}\\
= {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c\\
= {\log _{abc}}abc\\
= 1\left( {proved} \right)
\end{array}[/tex]
