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Prove that if a and b are the lengths of the legs and c is the length of hypotenuse of a right angle triangle, c-b not equal to 1; c+b not equal to 1 then $log_{(c+b)}a+log_{(c-b)}a=2log_{(c+b)}alog_{(c-b)}a$

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Where are you getting these problems? –  Will Jagy Feb 18 '13 at 20:08

1 Answer 1

From the definition of log, this equation is the same as $\frac{\log a}{\log{c+b} } +\frac{\log a}{\log{c-b}} =2\frac{\log a}{\log{c+b} }\frac{\log a}{\log{c-b}} $.

Canceling $\log a$ and clearing fractions, this becomes $\log(c-b) + \log(c+b) = 2 \log a$ or $(c-b)(c+b) = a^2$ or $c^2-b^2 = a^2$ which is just Pyth's theorem.

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thanks a lot ..that was great explanation. –  Sachin Sharmaa Feb 18 '13 at 7:04

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