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Math Help - Transposition....

  1. #1
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    Transposition....

    Hi I am new to this forum and require some help on transposition.... B is required to be the subject and the formula is

    C = B x log28/ 1 + A

    Any help would be great
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  2. #2
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    re: Transposition....

    Quote Originally Posted by red9 View Post
    Hi I am new to this forum and require some help on transposition.... B is required to be the subject and the formula is

    C = B x log28/ 1 + A

    Any help would be great
    if you meant ...

    C = B \cdot \frac{\log_2{8}}{1+A}

    C = B \cdot \frac{3}{1+A}

    \frac{C(1+A)}{3} = B
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  3. #3
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    re: Transposition....

    Quote Originally Posted by red9 View Post
    Hi I am new to this forum and require some help on transposition.... B is required to be the subject and the formula is

    C = B x log[SIZE=1]2[SIZE=3]8/ 1 + A
    Why don't you use grouping symbols.
    Do you mean C=\frac{Bx\log_2(8)}{1+A} OR C=\frac{Bx\log_2(8)}{1}+A~?

    Or something altogether different?
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  4. #4
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    re: Transposition....

    Hi Skeeter thanks very much, transposition is bit of a downfall for me but Im sure if I keep trying it will sink in...
    Thanks again
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  5. #5
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    Re: Transposition....

    Quote Originally Posted by red9 View Post
    C = B x log[SIZE=1]2[SIZE=3]8/ 1 + A
    Please use * as multiplication sign; also, brackets required: (1 + A)

    As a suggestion, seems to be nothing preventing simplification:

    Let k = log(2)8; your equation becomes:
    C = B*k / (1 + A)
    cross multiply:
    Bk = C (1 + A)
    divide by k:
    B = C(1 + A) / k
    Last edited by Wilmer; March 7th 2012 at 07:39 AM.
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  6. #6
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    Re: Transposition....

    Thank you Wilmer,
    How about to get A as te subject from the orignal equation?
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  7. #7
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    Re: Transposition....

    NOTE: edited my previous post...had a "watching hockey game" error!
    Quote Originally Posted by red9 View Post
    How about to get A as te subject from the orignal equation?
    original: C = B*k / (1 + A)
    cross multiply:
    C(1 + A) = Bk
    C + AC = Bk
    AC = Bk - C
    A = (Bk - C) / C

    I'll trust that you're "following/understanding" this, not just using as homework answer...

    If you come back with a similar problem......SHOW your work, else I'm busy!
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