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Math Help - Can i write this in another way? using Ln instead of Log

  1. #1
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    Can i write this in another way? using Ln instead of Log

    4/243Log(3x-2)-4/243log(3x+2)
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  2. #2
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    Quote Originally Posted by aa751 View Post
    4/243Log(3x-2)-4/243log(3x-2)
    Are my eyes playing tricks on me?

    \frac{4}{243}\log (3x-2)-\frac{4}{243}\log (3x-2)

    If so the answer is 0.

    What base is the log here? If you want to use ln you can use the change of base formula. It is \log_ab = \frac{\ln b}{\ln a}
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  3. #3
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    OH am sorry slides, It is my eyes that isnt working it actually looks like this

    4/243Log(3x-2)-4/243log(3x+2)
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  4. #4
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    Maybe start by taking out a common factor here

    \frac{4}{243}\log (3x-2)-\frac{4}{243}\log (3x+2)

    \frac{4}{243}\left(\log (3x-2)-\log (3x+2)\right)

    \frac{4}{243}\left(\log \frac{(3x-2)}{(3x+2)}\right)

    Depending on what you want to do next you can now use the change of base formula mentioned in post #2. What base is your logarithm?
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    I am not sure =//// I do not really know All i know is that the question wants me to convert that into Ln of f(x)/g(x) I am not sure about the base but I think its 10 It is prolly to the base of 10, that makes most sense right now
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  6. #6
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    THe log is to the base 10
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    \frac{4}{243}\left(\log_{10}\frac{(3x-2)}{(3x+2)}\right)

    \frac{4}{243}\left(\frac{\ln\frac{(3x-2)}{(3x+2)}}{\ln 10}\right)
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  8. #8
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    Ahh thanks man you are my hero , you saved me
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  9. #9
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    Quote Originally Posted by aa751 View Post
    (4/243)Log(3x-2)-(4/243)log(3x+2) Base of Log= 10 the answer should look like something like this Ln(g(x)/h(x))
    difference property of logs ...

    \log(x) - \log(y) = \log\left(\frac{x}{y}\right)


    change of base ...

    \log_b(a) = \frac{\ln{a}}{\ln{b}}
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