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Math Help - Logarithms..

  1. #1
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    Exclamation Logarithms..

    log(10 + 10^1/3)

    ^ = Raised to (Exponentiation)

    Another few should be coming your way. Please, I need help urgently!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ruler of Hell View Post
    log(10 + 10^1/3)

    ^ = Raised to (Exponentiation)

    Another few should be coming your way. Please, I need help urgently!
    Unless you are going to approximate it in some way, there's nothing you can do with it.
    log(a + b)
    has no expanded form in terms of a and b.

    (Note that log(ab) = log(a) + log(b).)

    -Dan
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  3. #3
    Bar0n janvdl's Avatar
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    Well its log(10^1 + 10^1/3)
    Cant we get the 1 and the 1/3 out of there somehow? However, it doesnt really seem likely to me.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ruler of Hell View Post
    log(10 + 10^1/3)

    ^ = Raised to (Exponentiation)

    Another few should be coming your way. Please, I need help urgently!
    i was going to make the same comment as topsquark. there is no nice algebraic way to break this down, so unless you can use log tables or a calculator (which i don't think is the idea), it's near impossible to do this. are you sure it shouldn't be 10*10^(1/3)
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Jhevon View Post
    i was going to make the same comment as topsquark. there is no nice algebraic way to break this down, so unless you can use log tables or a calculator (which i don't think is the idea), it's near impossible to do this. are you sure it shouldn't be 10*10^(1/3)
    Yes, are you sure you didnt make a typo?

    If it should be 10*10^(1/3) then
    log(10 x 10^1/3) = log(10) + log(10^1/3)
    = 1 + 1/3
    = 4/3
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  6. #6
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    That's what I was thinking. It has to be a mistake with the sum, I suppose.


    PROVE
    log4 + log2 = log(to the base 3)9
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Ruler of Hell View Post
    That's what I was thinking. It has to be a mistake with the sum, I suppose.


    PROVE
    log4 + log2 = log(to the base 3)9
    RHS: log_3 9 = 2

    LHS: log_10 4 + log_10 2 = log_10 8 = 0,9
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  8. #8
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    I didn't understand you. I got RHS as 2, and I got LHS as 3log2. Umm... How is that proved?
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ruler of Hell View Post
    That's what I was thinking. It has to be a mistake with the sum, I suppose.


    PROVE
    log4 + log2 = log(to the base 3)9
    as you can probably tell from janvdl's answer, this is not true. maybe the book made another typo
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  10. #10
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Ruler of Hell View Post
    I didn't understand you. I got RHS as 2, and I got LHS as 3log2. Umm... How is that proved?
    3log_10 2 = log_10 8
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  11. #11
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    Not a typo, incorrect sums given in the book. Thanks for the help!
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ruler of Hell View Post
    Not a typo, incorrect sums given in the book. Thanks for the help!
    i'm seriously considering telling you to through that book away but if that's what your school uses, i guess you can't do that

    books make mistakes once in a while, but usually not often enough for you to pin point two in the same section, and such blatant mistakes to top it all off
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ruler of Hell View Post
    That's what I was thinking. It has to be a mistake with the sum, I suppose.


    PROVE
    log4 + log2 = log(to the base 3)9
    Please start a new thread for a new problem.

    If no base is given log(4) can be either of two things, depending on who is teaching the class and what field you are in. It could either be
    log_{10}(4)
    or
    log_e(4) = ln(e)

    However as log_3(9) = log_3(3^2) = 2 I can think of only one way such that this is going to work:

    log_b(4) + log_b(2) = 2

    log_b(4*2) = 2

    log_b(8) = 2

    Using the change of base formula:
    log_b(8) = ln(8)/ln(b) = 2

    ln(b) = (1/2)ln(8)

    b = e^{(1/2)ln(8)} = [e^{ln(8)}]^{1/2} = sqrt{8} = 2.82843

    So we CAN say:
    log_{sqrt{8}}(4) + log_{sqrt{8}}(2) = log_3(9)

    but I doubt this is what the question was after.

    -Dan
    Last edited by topsquark; May 8th 2007 at 07:00 AM.
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