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Math Help - Right angle triangle - logarithm problem

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    Right angle triangle - logarithm problem

    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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  2. #2
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    Re: Right angle triangle - logarithm problem

    Quote Originally Posted by sachinrajsharma View Post
    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
    1. Use the base-change-formula: \log_b(a)=\frac{\ln(a)}{\ln(b)}

    2. Your equation becomes:

    log_{c+b}a+log_{c-b}a=2log_{c+b}alog_{c-b}a~\implies~ \\ \\ \frac{\ln(a)}{\ln(c+b)}+\frac{\ln(a)}{\ln(c-b)}= 2 \cdot \frac{\ln(a)}{\ln(c+b)} \cdot \frac{\ln(a)}{\ln(c-b)}

    3. Divide through by ln(a) and multiply through by \ln(c+b) \cdot \ln(c-b) . You'll get:

    \ln(c-b) + \ln(c+b) = 2\ln(a)

    Use laws of logarithms. You'll get:

    (c-b)(c+b)=a^2

    Expand and you'll get the Pythagorean theorem.
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