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    logb + log c = 3

    find b in terms of c

    I just dont know, I get to log b = 3 - logc

    then I seem to get to b = 10^3-logc

    then b = 1000 x 1/logc

    the answer gets rid of the log and is just 1000/c

    why is this?

    ^ This has been done, I have another question at the bottom Please help!
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    You know that loga + logb = log(ab)
    Can you change 3 into a log?
    If you can then you can delog both sides.
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    (Original post by 3607)
    logb + log c = 3

    find b in terms of c

    I just dont know, I get to log b = 3 - logc

    then I seem to get to b = 10^3-logc


    then b = 1000 x 1/logc


    the answer gets rid of the log and is just 1000/c

    why is this?
    The bold's your problem.
    \log b = 3 -\log c 

\Rightarrow \log b = 10^{3-\log c}
    Remember that 10^{\log_{10}f(x)} = f(x)

    EDIT: The above poster is using my preferred method though this should still work
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    (Original post by maggiehodgson)
    If you can then you can delog both sides.
    I think the preferred term might be to take the antilog of both sides.

    "delog" smacks of deforestation of the jungle and enviromental concerns. +rep!
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    Thanks.

    I'm not very technical am I. What does that +rep mean?
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    (Original post by maggiehodgson)
    Thanks.

    I'm not very technical am I. What does that +rep mean?
    It means I gave your post a "postive reputation". The "1" next to the thumbs up.

    My comment was just an amusing aside, not to be taken with any great seriousness.
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    (Original post by joostan)
    The bold's your problem.
    \log b = 3 -\log c 

\Rightarrow \log b = 10^{3-\log c}
    Remember that 10^{\log_{10}f(x)} = f(x)</b>

    EDIT: The above poster is using my preferred method though this should still work

    Ah, thanks, I don't think I've ever been taught that haha.

    Thanks to everyone for trying to help, I found this the most helpful and easy to get
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    (Original post by 3607)
    Ah, thanks, I don't think I've ever been taught that haha.

    Thanks to everyone for trying to help, I found this the most helpful and easy to get
    Lol, no prob
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    (Original post by ghostwalker)
    It means I gave your post a "postive reputation". The "1" next to the thumbs up.

    My comment was just an amusing aside, not to be taken with any great seriousness.

    Thanks for the vote. I'm pretty new to TSR so am not with what can and can't be done.

    Also, I didn't really take your comment as a chastisement You've helped me a lot with some mechanics study so I know you're a kind sort of person.

    Will be back on mechanics as soon as S1B is out of the way.
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    (Original post by joostan)
    The bold's your problem.
    \log b = 3 -\log c 

\Rightarrow \log b = 10^{3-\log c}
    Remember that 10^{\log_{10}f(x)} = f(x)

    EDIT: The above poster is using my preferred method though this should still work
    Isn't it b = 10^(3-log c)?
    then you cancel the log from the log c because it's log to the base 10 as a power of 10?
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    (Original post by joostan)
    The bold's your problem.
    \log b = 3 -\log c 

\Rightarrow \log b = 10^{3-\log c}
    Remember that 10^{\log_{10}f(x)} = f(x)

    EDIT: The above poster is using my preferred method though this should still work
    I've got another question for everyone, it's not absolutely vital this is answered, but I would be interested to know why the above is true!

    Thanks to anyone who helps
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    (Original post by 3607)
    I've got another question for everyone, it's not absolutely vital this is answered, but I would be interested to know why the above is true!

    Thanks to anyone who helps
    The 10 to the power of is the inverse operater of the log. A function and it's inverse acting on something yields that something.
    The definition is that f^{-1}(f(x)) = x
 
 
 
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