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Math Help - Logarithm bases

  1. #1
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    Logarithm bases

    Im very confused with problem and i hope you can help me.


    How to calculate the nth term of the fallowing sequences, giving the answer in the form p/q, where p, q ∈ Z

    log_{1/5}(125) , log_{1/125}(125) , log_{1/625}(125)

    log_{4}(64) , log_{8}(x) , log_{32}(64)




    Also describe how to obtain the third answer in each row from the first two answers.
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  2. #2
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    Quote Originally Posted by erickvm123 View Post
    Im very confused with problem and i hope you can help me.


    How to calculate the nth term of the fallowing sequences, giving the answer in the form p/q, where p, q ∈ Z

    log_{1/5}(125) , log_{1/125}(125) , log_{1/625}(125)
    Each of these terms is a logarithm of 125 to bases of 1/5, 1/125= 1/5^3, and 1/625= 1/5^4. Frankly, I don't see a pattern unless there is a missing log_{1/25}(125) in there. If that were the case, then the nth term would be log_{1/5^n}(125).

    The whole point of a logarithm is that log_a(a^x)= x. If the nth term is indeed log_{1/5^n}(125), then 125= 5^3= (5^n)^{3/n}= (1/5^n)^{-3/n}. The nth term would be -\frac{3}{n}.

    log_{4}(64) , log_{8}(x) , log_{32}(64)




    Also describe how to obtain the third answer in each row from the first two answers.
    log_8(x)?? Surely, that is wrong. It must be log_8(64) like the others. In that case, each is the logarithm of 64 to bases 4= 2^2, 8= 2^3, and 16= 2^4. The nth term would be log_{2^{n-1}}(64). 64= 8^2= (2^3)^2= 2^6= (2^{n-1})^{6/(n-1)}. log_{2^{n-1}}(64)= \frac{6}{n-1}.
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  3. #3
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    Hello, erickvm123!

    I agree with HallsofIvy . . . the changing bases have no patten.
    Besides the typos, I believe you left out some terms . . .


    Calculate the n^{th} term of the fallowing sequences,
    giving the answer in the form \tfrac{p}{q} where p,q \in Z

    \log_{4}(64),\quad \log_{8}(64), \quad \log_{16}(64),\quad \log_{32}(64),\quad \hdots

    Evaluating the logs, we have: . 3,\;2,\;\frac{3}{2},\;\frac{6}{5},\;\hdots

    Then we see the patten: . \frac{6}{2},\;\frac{6}{3},\;\frac{6}{4},\;\frac{6}  {5},\;\hdots

    Therefore, the n^{th} term is: . \frac{6}{n+1}

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  4. #4
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    well this is how it is in my paper that's why I'm confused, but thank you for finding the general term in p/q. And yeah it should be log_{8} 64 instead of log_{8} (X)
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