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Math Help - Logistic Problem - Help Please!

  1. #1
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    Logistic Problem - Help Please!

    'How can you get "very close" to Log(7) if you knew Log(2) and Log(3)?


    I'm still wondering about this; if anyone gets close.. just tell me and I'll tweak your answer more. Thanks in advance!
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by .Masta X View Post
    'How can you get "very close" to Log(7) if you knew Log(2) and Log(3)?


    I'm still wondering about this; if anyone gets close.. just tell me and I'll tweak your answer more. Thanks in advance!

    Hint: Remember that
    \log(xy)=\log(x)+\log(y)
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  3. #3
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    Hello, Masta X!

    How can you get "very close" to Log(7) if you knew Log(2) and Log(3)?
    Is this close enough?


    We have: . \frac{2^{21}}{3\cdot10^5} \;=\;\frac{2,\!097,\!152}{300,\!000} \;=\;<br />
6.990506667


    Therefore: . 21\log2 - \log 3 - 5 \;=\;0.844508654  \;\approx\;\log 7

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  4. #4
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    Yes, Soroban!
    I got something like that, but I was off by 0.0020.


    Thank you very much! I really appreciate your help greatly!
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