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  1. #1
    Lil
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    I need without calculator improve that: \frac{1}{log_35}+\frac{1}{log_95}>2

    what I need to do with log_?5 ?
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  2. #2
    A Plied Mathematician
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    I would try to write both logarithms using the same base. It's very awkward using different bases. How could you do that?
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  3. #3
    Lil
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    Use this formula: log_am=\frac{logm}{loga}?
    so, \frac{log5}{log3} and \frac{1}{2}\frac{log5}{log3}

    it's this or not?
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  4. #4
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    You're correct, but you can go further. I would agree that

    \log_{9}(5)=\dfrac{\ln(5)}{\ln(9)}=\dfrac{\ln(5)}{  \ln(3^{2})}=\dfrac{\ln(5)}{2\ln(3)}=\dfrac{1}{2}\l  og_{3}(5).

    So now what can you do?
    Last edited by Ackbeet; November 6th 2010 at 03:12 AM. Reason: Lil correct.
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  5. #5
    Lil
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    Quote Originally Posted by Ackbeet View Post
    You're correct, but you can go further. I would agree that

    \log_{9}(5)=\dfrac{\ln(5)}{\ln(9)}=\dfrac{\ln(5)}{  \ln(3^{2})}=\dfrac{\ln(5)}{2\ln(3)}=\dfrac{1}{2}\l  og_{3}(5).
    with this =log_3\sqrt{5}
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  6. #6
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    You can do that, but I'm not sure I'd recommend it. The reason is that if you leave it at \dfrac{1}{2}\log_{3}(5), then it's easier to get a common denominator on the LHS of the original expression.

    So, take a look at the original expression, and substitute in this \dfrac{1}{2}\log_{3}(5). What do you get?
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  7. #7
    Lil
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    So, I have this \frac{1}{log\frac{5}{3}}+\frac{1}{\frac{1}{2}log_3  5}>2
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  8. #8
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    Not quite. I would write this:

    \dfrac{1}{\log_{3}(5)}+\dfrac{1}{\dfrac{1}{2}\log_  {3}(5)}>2.

    (No need to change the first fraction.) So that is what we need to prove. What would you do next?
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  9. #9
    Lil
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    Can I use: \frac{1}{log_ma}=log_am
    then log_53+\frac{1}{2}log_53>2
    or
    log_53+log_59>log_525
    log_527>log_525
    27>25
    Last edited by Lil; November 9th 2010 at 07:58 AM.
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  10. #10
    A Plied Mathematician
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    Sure you can. However, you would get

    \log_{5}(3)+2\log_{5}(3)>2, not what you got.
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