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Math Help - Couple of questions about logarithms and exponents

  1. #1
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    Couple of questions about logarithms and exponents

    Which is the proper way of condensing this:

    ln2-4ln3

    Is it:

    ln2+ln3^(-4)

    or

    ln2-ln3^4

    Are either of these ways valid? And if so, which is the appropriate way? Or does that matter on the problem?


    My other question.. Are these valid:
    3^(x+1) = (3^x)*3
    3^(x-1) = (3^x)/x or (3^x)(3^-1)
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  2. #2
    Super Member Gamma's Avatar
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    ln2-4ln3=ln2-ln3^4=\frac{ln2}{ln3^4}

    3^{x+1}=3^x3^1=3(3^x)

    3^{x-1}=3^x3^{-1}=\frac{1}{3}(3^x)
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  3. #3
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    Quote Originally Posted by Phire View Post
    Which is the proper way of condensing this:

    ln2-4ln3

    Is it:

    ln2+ln3^(-4)

    or

    ln2-ln3^4

    Are either of these ways valid? And if so, which is the appropriate way? Or does that matter on the problem?
    ln2+ ln 3^(-4)= ln[2(3^-4)]= ln(2/81)
    ln2- ln 3^4= ln[2/3^4]= ln(2/81)

    They are exactly the same.


    My other question.. Are these valid:
    3^(x+1) = (3^x)*3
    Yes, 3^(x+1)= 3^x*3^1= (3^x)*3

    3^(x-1) = (3^x)/x or (3^x)(3^-1)
    3^(x-1)= (3^x)(3^-1)= (3^x)/3. Is that what you meant instead of (3^x)/x?
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Gamma View Post
    ln2-4ln3=ln2-ln3^4=\frac{ln2}{ln3^4}

    3^{x+1}=3^x3^1=3(3^x)

    3^{x-1}=3^x3^{-1}=\frac{1}{3}(3^x)
    ln(a)-ln(b) \neq \frac{ln(a)}{ln(b)}

    Test with a = e and b=1

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