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Math Help - Natural logarithms

  1. #1
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    Natural logarithms

    1) ln( x^{\frac{1}{2}})=2.
    We know: ln( x^{n})=n ln(x)
    -> 2ln(x)=ln(2) --> ln(x)= {\frac{ln(2)}{2}}. --> x= 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:
    \frac {1}{2}*ln(x)=2

    Multiple by 2

    ln(x)=4

    Raise to the e

    x={e}^{4}
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  3. #3
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    Quote Originally Posted by Schdero View Post
    1) ln( x^{\frac{1}{2}})=2.
    We know: ln( x^{n})=n ln(x)
    -> 2ln(x)=ln(2) --> ln(x)= {\frac{ln(2)}{2}}. --> x= 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  \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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    ln(x^{2})=2

    2ln(x)=2

    ln(x)=1

    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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