উদাহরণ ৯৷[tex]a > 0,c > b = \sqrt {ac} ,a,c,ac \ne 1,N > 0[/tex] হলে প্রমান করো
[tex]\frac{{{{\log }_a}N}}{{{{\log }_c}N}} = \frac{{{{\log }_a}N - {{\log }_b}N}}{{{{\log }_b}N - {{\log }_c}N}}[/tex] [Jt.Ent.’88]
প্রমান:
মনে করি
[tex]\begin{array}{l}
{\log _a}N = x \Rightarrow N = {a^x} \Rightarrow a = {N^{\frac{1}{x}}} \to \left( 1 \right)\\
{\log _b}N = y \Rightarrow N = {b^y} \Rightarrow b = {N^{\frac{1}{y}}} \to \left( 2 \right)\\
{\log _c}N = z \Rightarrow N = {c^z} \Rightarrow c = {N^{\frac{1}{z}}} \to \left( 3 \right)
\end{array}[/tex]
(1), (2) ও (3) থেকে পাই
[tex]{a^x} = {b^y} = {c^z} \to \left( 4 \right)[/tex]
শর্তানুযায়ী [tex]b = \sqrt {ac} [/tex][ (1), (2), (3) ]থেকে পাই
[tex]\begin{array}{l}
\Rightarrow {N^{\frac{1}{y}}} = \sqrt {{N^{\frac{1}{x}}}{N^{\frac{1}{z}}}} \\
\Rightarrow {N^{\frac{2}{y}}} = {N^{\frac{1}{x}}}{N^{\frac{1}{z}}}\\
\Rightarrow {N^{\frac{2}{y}}} = {N^{\frac{1}{x} + \frac{1}{z}}}\\
\Rightarrow \frac{2}{y} = \frac{1}{x} + \frac{1}{z}\\
\Rightarrow \frac{1}{y} - \frac{1}{z} = \frac{1}{x} - \frac{1}{y}\\
\Rightarrow \frac{{z - y}}{{yz}} = \frac{{y - x}}{{xy}}\\
\Rightarrow \frac{{y - z}}{z} = \frac{{x - y}}{x}\\
\Rightarrow \frac{x}{z} = \frac{{x - y}}{{y - z}}\\
\Rightarrow \frac{{{{\log }_a}N}}{{{{\log }_c}N}} = \frac{{{{\log }_a}N - {{\log }_b}N}}{{{{\log }_b}N - {{\log }_c}N}}\left( {proved} \right)
\end{array}[/tex]