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Prove that if $a$ and $b$ are the lengths of the legs and $c$ is the length of the hypotenuse of a right angle triangle, $c-b \neq 1$, $c+b \neq 1$ then $\log_{(c+b)}a+\log_{(c-b)}a=2\log_{(c+b)}a\log_{(c-b)}a$.

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

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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