Problem 009 | Logarithm | bengalstudents.com

Bengalstudents ...Sun, 17/02/2013 - 9:00pm

উদাহরণ ৯৷ $a > 0,c > b = \sqrt {ac} ,a,c,ac \ne 1,N > 0$ হলে প্রমান করো

$\frac{{{{\log }_a}N}}{{{{\log }_c}N}} = \frac{{{{\log }_a}N - {{\log }_b}N}}{{{{\log }_b}N - {{\log }_c}N}}$ [Jt.Ent.’88]

প্রমান:

মনে করি

$\begin{array}{l} {\log _a}N = x \Rightarrow N = {a^x} \Rightarrow a = {N^{\frac{1}{x}}} \to \left( 1 \right)\\ {\log _b}N = y \Rightarrow N = {b^y} \Rightarrow b = {N^{\frac{1}{y}}} \to \left( 2 \right)\\ {\log _c}N = z \Rightarrow N = {c^z} \Rightarrow c = {N^{\frac{1}{z}}} \to \left( 3 \right) \end{array}$

(1), (2) ও (3) থেকে পাই

${a^x} = {b^y} = {c^z} \to \left( 4 \right)$

শর্তানুযায়ী $b = \sqrt {ac}$ [ (1), (2), (3) ]থেকে পাই

$\begin{array}{l} \Rightarrow {N^{\frac{1}{y}}} = \sqrt {{N^{\frac{1}{x}}}{N^{\frac{1}{z}}}} \\ \Rightarrow {N^{\frac{2}{y}}} = {N^{\frac{1}{x}}}{N^{\frac{1}{z}}}\\ \Rightarrow {N^{\frac{2}{y}}} = {N^{\frac{1}{x} + \frac{1}{z}}}\\ \Rightarrow \frac{2}{y} = \frac{1}{x} + \frac{1}{z}\\ \Rightarrow \frac{1}{y} - \frac{1}{z} = \frac{1}{x} - \frac{1}{y}\\ \Rightarrow \frac{{z - y}}{{yz}} = \frac{{y - x}}{{xy}}\\ \Rightarrow \frac{{y - z}}{z} = \frac{{x - y}}{x}\\ \Rightarrow \frac{x}{z} = \frac{{x - y}}{{y - z}}\\ \Rightarrow \frac{{{{\log }_a}N}}{{{{\log }_c}N}} = \frac{{{{\log }_a}N - {{\log }_b}N}}{{{{\log }_b}N - {{\log }_c}N}}\left( {proved} \right) \end{array}$

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Problem 009 | Logarithm

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