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Math Help - Logarithms. Please help.

  1. #1
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    Logarithms. Please help.

    Expanding:

    Log(base 5)( radical x-1/x+1 )=

    Combining the expression:

    Log(base 5)(x^2 - 1) - Log(base 5)(x - 1) =

    Simplify:

    (Log(base 2)5)(Log(base 5)7) =
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  2. #2
    MHF Contributor
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    Quote Originally Posted by aligator0207 View Post
    Expanding:

    Log(base 5)( radical x-1/x+1 )=

    Combining the expression:

    Log(base 5)(x^2 - 1) - Log(base 5)(x - 1) =

    Simplify:

    (Log(base 2)5)(Log(base 5)7) =
    hi

    for (1) , is it \log_{5}(\sqrt{\frac{x-1}{x+1}}) ?

    (2) \log_{2}5\cdot \log_{5}7=\frac{\log_{2}5}{\log_{2}2}\cdot \frac{\log_{2}7}{\log_{2}5}

     <br />
=\log_{2}7<br />
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  3. #3
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    Yes, that's how you write out number 1. Do you know how to do number 2?
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  4. #4
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    Hello, aligator0207!

    Expand: . \log_5 \sqrt{\frac{x-1}{x+1}}

    \log_5\sqrt{\frac{x-1}{x+1}} \;=\;\log_5\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}} \;=\;\tfrac{1}{2}\log_5\left(\frac{x-1}{x+1}\right) \;=\;\tfrac{1}{2}\bigg[\log_5(x-1) - \log_5(x+1)\bigg]




    Combine: . \log_5(x^2 - 1) - \log_5(x - 1)

    \log_5(x^2-1) - \log_5(x-1) \;=\;\log_5\left(\frac{x^2-1}{x-1}\right) \;=\;\log_5\left(\frac{(x-1)(x+1)}{x-1}\right) \;=\;\log_5(x+1)




    Simplify: . \log_2(5)\cdot\log_5(7)
    Base-change Formula: . \log_b(a) \:=\:\frac{\log (a)}{\log(b)}


    \log_2(5)\cdot\log_5(7) \;=\;\frac{\log(5)}{\log(2)} \cdot\frac{\log(7)}{\log(5)} \;=\;\frac{\log(7)}{\log(2)} \;=\;\log_2(7)

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  5. #5
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    Sweet! Thank you so much. At least I got the first two right
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