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Thread: ln(1/2) & ln(e^k^t)

  1. #1
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    ln(1/2) & ln(e^k^t)

    $\displaystyle \frac{1}{2}=e^k^t$

    When ln() is taken to both sides of the equation it is supposed to simplify to:

    $\displaystyle -ln(2)=kt$

    I'm wondering by what rules was each side simplified. I know that ln(e) is 1 but in this case ln(e^k^t) is the quantity k*t? The fact that ln(1/2) simplifies somehow to -ln(2) just leaves me bewildered at this point.

    Thanks in advance for any posts on this...
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  2. #2
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    Re: ln(1/2) & ln(e^k^t)

    Quote Originally Posted by sepoto View Post
    $\displaystyle \frac{1}{2}=e^k^t$

    When ln() is taken to both sides of the equation it is supposed to simplify to:

    $\displaystyle -ln(2)=kt$

    I'm wondering by what rules was each side simplified. I know that ln(e) is 1 but in this case ln(e^k^t) is the quantity k*t? The fact that ln(1/2) simplifies somehow to -ln(2) just leaves me bewildered at this point.

    Thanks in advance for any posts on this...
    You know that
    $\displaystyle ln(a^n) = n \cdot ln(a)$

    So write $\displaystyle \frac{1}{2} = 2^{-1}$. So what is $\displaystyle ln(2^{-1})$?

    -Dan

    Edit: As to your other question, the RHS is $\displaystyle e^{kt}$, not e^{k^t}. So $\displaystyle ln(e^{kt}) = kt \cdot ln(e) = kt$.
    Last edited by topsquark; Nov 18th 2013 at 12:48 PM.
    Thanks from sepoto
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  3. #3
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    Re: ln(1/2) & ln(e^k^t)

    Another way to look at ln(1/2): ln(A/B)= ln(A)- ln(B) so ln(1/2)= ln(1)- ln(2)= 0- ln(2)= -ln(2).
    Thanks from sepoto and topsquark
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