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Math Help - value of log

  1. #1
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    value of log

    (a) e ^ -2ln5

    (b)ln(ln e^e^10)

    The second one is e to the e to the tenth power.
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  2. #2
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    Quote Originally Posted by dm10 View Post
    (a) e ^ -2ln5

    (b)ln(ln e^e^10)

    The second one is e to the e to the tenth power.
    For a) Do you remember the rule \log_a(m^p)=p\log_a(m)?

    Using this rule, we get

    e^{-2\ln(5)}=e^{\ln(5^{-2})}=5^{-2}=\frac{1}{5^2}=\frac{1}{25}

    The second actually just reduces to 10, because \ln (e^{e^{10}})=e^{10} and so
    \ln[\ln(e^{e^{10}})]=\ln[e^{10}]=10.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    For a) Do you remember the rule \log_a(m^p)=p\log_a(m)?

    Using this rule, we get

    e^{-2\ln(5)}=e^{\ln(5^{-2})}=5^{-2}=\frac{1}{5^2}=\frac{1}{25}

    The second actually just reduces to 10, because \ln(e^e^10)=e^10 and so
    \ln[\ln(e^e^10)]=\ln[e^10]=10.
    Thank you, you're answers really help me understand the material.
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  4. #4
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    Quote Originally Posted by dm10 View Post
    Thank you, you're answers really help me understand the material.
    No problem, and I just fixed the LaTeX errors in my post.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    No problem, and I just fixed the LaTeX errors in my post.
    I've got one more left if you don't mind helping:

    arcsin 2

    Find the exact value.
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  6. #6
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    Quote Originally Posted by dm10 View Post
    I've got one more left if you don't mind helping:

    arcsin 2

    Find the exact value.
    Do you realise that arcsin 2 has no real value?
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Do you realise that arcsin 2 has no real value?
    So the answer would be no real value?
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  8. #8
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    Quote Originally Posted by dm10 View Post
    So the answer would be no real value?
    If you're working exclusively with real numbers it would be reasonable to say that arcsin(2) is not defined.
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