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Math Help - rewriting a formula

  1. #1
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    Question rewriting a formula

    Here's the original formula.

    B = log (X/Y)

    I have to rewrite it to be Y = (something). I tried to do it but I'm not real sure I did it right.

    I thought I could unlog it first

    10^B = X/Y

    Then divide by Y to get Y on the other side.

    10^B / Y = X

    Then I turned it over.

    Y / 10^B = 1 / X

    Then divide both sides by 10^B.

    Y = 1 / ( X / 10^B )

    Am I doing it right? Or close to right?
    Raven
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Raven View Post
    ...

    Y / 10^B = 1 / X

    Then divide both sides by 10^B.

    Y = 1 / ( X / 10^B )

    Am I doing it right? Or close to right?
    Raven
    isn't the 10^B already dividing the left? we need to multiply both sides by 10^B
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  3. #3
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    Question

    So from

    Y / 10^B = 1 / X

    it's now

    Y = (10^B)(1/X)

    and from that

    Y = 10^B / X

    ok?
    Raven
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Raven View Post
    So from

    Y / 10^B = 1 / X

    it's now

    Y = (10^B)(1/X)

    and from that

    Y = 10^B / X

    ok?
    Raven
    nope.

    you made the same error with Y earlier, i just didn't catch it until now. you should have multiplied both sides by Y when you divided
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  5. #5
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    Question

    ok, do over

    B = log (X/Y)
    10^B = X/Y
    (10^B)Y = X
    Y = X / 10^B
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Raven View Post
    ok, do over

    B = log (X/Y)
    10^B = X/Y
    (10^B)Y = X
    Y = X / 10^B
    yes, correct


    An alternate solution:

    B = \log \left( \frac XY \right)

    \Rightarrow B = \log X - \log Y

    \Rightarrow \log Y = \log X - B

    \Rightarrow Y = 10^{\log X - B}

    \Rightarrow Y = 10^{\log X} \cdot 10^{-B}

    \Rightarrow Y = 10^{-B}X
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