Problem 009 | Logarithm

Submitted by Anonymous (not verified) on Sun, 02/17/2013 - 21:00

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]

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