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Thread: Natural logarithms

  1. #1
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    Natural logarithms

    1) ln($\displaystyle x^{\frac{1}{2}}$)=2.
    We know: ln($\displaystyle x^{n}$)=n ln(x)
    -> 2ln(x)=ln(2) --> ln(x)=$\displaystyle {\frac{ln(2)}{2}}$. --> x= $\displaystyle e^{\frac{ln(2)}{2}}$.
    Is that correct? Because my booklet, again, says something different

    2) are generally all logarithmic rules appliable for ln?
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  2. #2
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    I obtained a this:
    $\displaystyle \frac {1}{2}*ln(x)=2$

    Multiple by 2

    $\displaystyle ln(x)=4$

    Raise to the e

    $\displaystyle x={e}^{4}$
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  3. #3
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    Quote Originally Posted by Schdero View Post
    1) ln($\displaystyle x^{\frac{1}{2}}$)=2.
    We know: ln($\displaystyle x^{n}$)=n ln(x)
    -> 2ln(x)=ln(2) --> ln(x)=$\displaystyle {\frac{ln(2)}{2}}$. --> x= $\displaystyle e^{\frac{ln(2)}{2}}$.
    Is that correct? Because my booklet, again, says something different

    2) are generally all logarithmic rules appliable for ln?
    1. answered above.

    2. yes because ln(x) is the same thing as $\displaystyle \log_e(x) $
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  4. #4
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    Yes sure, sry i meant ln(x^2)=2, not x^1/2. But thats no big change, because in my booklet the solution is +/-e and this is nonsense, isn't it?
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  5. #5
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    $\displaystyle ln(x^{2})=2$

    $\displaystyle 2ln(x)=2$

    $\displaystyle ln(x)=1$

    $\displaystyle x=e$

    You can't have a negative value in the natural function since the domain is (0,infinity)

    -e=-2.78......
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