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Math Help - log identity

  1. #1
    Super Member bigwave's Avatar
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    log identity

    show that if {a}^{2}+{b}^{2}={c}^{2}, then

     {log}_{b+c}a + {log}_{c-b}a= 2\left( {log}_{b+c}a\right)\left( {log}_{c-b}a \right)

    I tried to solve this with change of base but didn't seem to help. so not sure even what the first step is on this one.
    Last edited by bigwave; May 4th 2011 at 02:17 PM. Reason: corrected base
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  2. #2
    Moo
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    Hello,

    Please show what you've tried
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  3. #3
    Super Member bigwave's Avatar
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    fyi i made a correction on the org post with the bases
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  4. #4
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    try {log}_{b+c}a^2+{log}_{c-b}b^2.

    hopefully you can work out what that's meant to be.
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  5. #5
    Super Member bigwave's Avatar
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    using change of base

    Quote Originally Posted by bigwave View Post
    show that if {a}^{2}+{b}^{2}={c}^{2}, then

     {log}_{b+c}a + {log}_{c-b}a= 2\left( {log}_{b+c}a\right)\left( {log}_{c-b}a \right)
    ok so change of base would be

     \frac{log(a)}{{log}(b+c)}} + \frac{log(a)}{{log}(c-b)}}= 2\left( {log}_{b+c}a\right)\left( {log}_{c-b}a \right)

    but how can you add these?
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  6. #6
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    Try changing all Logs to log_{a}. And remember log_{a}a=1.
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  7. #7
    Super Member bigwave's Avatar
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    Quote Originally Posted by poirot View Post
    try {log}_{b+c}a^2+{log}_{c-b}b^2.

    hopefully you can work out what that's meant to be.
    {log}_{b+c}a^2+{log}_{c-b}b^2.

    or

    2{log}_{b+c}a+2{log}_{c-b}b

    not sure where we are going with this.
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  8. #8
    Super Member bigwave's Avatar
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    Quote Originally Posted by VincentP View Post
    Try changing all Logs to log_{a}. And remember log_{a}a=1.
    what do you do with the coeficient 2
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  9. #9
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    sorry that b^2 should be another a^2 so 2(log(a)/log(b+c)+log(a)/log(c-b))and add the fractions. Then its a simple case of log laws. log(x)+log(y)=log(xy)
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  10. #10
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    After all replacements and multiplying with log_{a}(c-b) and log_{a}(c+b) you get this:
    log_{a}(c-b)+log_{a}(c+b) = 2
    Then you can apply the log law giving:
    log_{a}((c-b)(c+b)) = 2
    which simplifies to:
    log_{a}(c^2-b^2) = 2
    Now use the definition of logs to get this:
    a^2 = c^2 - b^2 Which is the same as c^2 = a^2 + b^2
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  11. #11
    Forum Admin topsquark's Avatar
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    Technically, you know, this should be the other way around. The question says that if a^2 + b^2 = c^2, so you need to also be sure you know how to reverse the argument.

    -Dan
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  12. #12
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    Thats what my method did topsquark
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by poirot View Post
    Thats what my method did topsquark
    Sorry, I forgot to quote. My comment was regarding VincentP's post. And, I think bigwave didn't get that point either.

    -Dan
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  14. #14
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    Doesn't the proof just show that there is a biconditional relation between the statements?
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  15. #15
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by VincentP View Post
    Doesn't the proof just show that there is a biconditional relation between the statements?
    It would take a better mind than mine to prove so without doing the work the other way. But my comment was that the problem statement said to start with a^2 + b^2 = c^2 and derive the log statement. So the answer should be in that form. So saying, when I did the problem I did it 'backward" like most people would, but then I reversed it to get the "correct" derivation. Nothing I could see tells me that going from the log statement to a^2 + b^2 = c^2 implied that the converse was true until I actually worked the steps.

    -Dan
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