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Math Help - Rules of logs?

  1. #1
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    Rules of logs?

    Why does ln(1/3) = -ln(3)? Is it because (1/3)^-1=3? And then you just put the -1 at the front? If this even makes sense... Thank you in advance.
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    Re: Rules of logs?

    Quote Originally Posted by Emanresu View Post
    Why does ln(1/3) = -ln(3)? Is it because (1/3)^-1=3? And then you just put the -1 at the front? If this even makes sense... Thank you in advance.
    Yes that's exactly the reason. One of the rules of logarithms states \displaystyle \begin{align*} \log{\left(m^p\right)} = p\log{m} \end{align*}, so

    \displaystyle \begin{align*} \ln{\left(\frac{1}{3}\right)} &= \ln{\left(3^{-1}\right)} \\ &= -1\ln{(3)} \\ &= -\ln{(3)} \end{align*}
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    Re: Rules of logs?

    You can show that ln(1/3) = -ln(3) in a couple of ways:

    1. As you noted, 1/3 = 3^(-1). Since ln(a^b) = b ln(a), if you let a = 3 and b = -1 you get ln(3^-1) = -1 ln(3)

    2. Recall that ln(a/b) = ln(a) - ln(b). If you let a = 1 and b = 3 you get ln(1/3) = ln(1)-ln(3) = 0-3 = -3

    Hope this helps.
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    Re: Rules of logs?

    Quote Originally Posted by Emanresu View Post
    Why does ln(1/3) = -ln(3)? Is it because (1/3)^-1=3? And then you just put the -1 at the front? If this even makes sense... Thank you in advance.
    That is true. But also

    \ln\left(\frac{1}{3}\right)=\ln(1)-\ln(3)=0-ln(3)=-\ln(3)
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    Re: Rules of logs?

    Thank you everyone, especially for that second method.
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