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Math Help - Logarithms and uncertainties: 0.434 and 2.303

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    Logarithms and uncertainties: 0.434 and 2.303

    I need to propagate a few uncertainties for logarithmic numbers, using the following two formulas:

    For: base 10         x = log_{10} a

    s_x=0.434(\frac{s_a}{a})

    where:
    s_x= uncertainty in result
    s_a= uncertainty in numbers used for calculation
    a= numbers used for calculation

    AND

    For: base 10         x = 10^a a

    s_x=2.303\cdot x\cdot s_a

    where:
    s_x= uncertainty in result
    x= result of calculation
    s_a= uncertainty in numbers used for calculation

    My question is, what exactly are the 0.434 and 2.303 values for? I assume they're to counter something that log calculations add?
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  2. #2
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    What do you mean by "propagate a few uncertainties?"
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    I'm actually using these formulas for chemistry (although they're aren't chemistry specific). In chemistry, or any other science, you have a margin of error in your results, and so you need to calculate the possible range of errors. These formulas are used to calculate the uncertainties in a value, ie. if my value is 60.0, then a possible range of uncertainties is +/- 0.5. I think the formulas are supposed to be "statistical" formulas?
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    Quote Originally Posted by maybealways View Post
    I'm actually using these formulas for chemistry (although they're aren't chemistry specific). In chemistry, or any other science, you have a margin of error in your results, and so you need to calculate the possible range of errors. These formulas are used to calculate the uncertainties in a value, ie. if my value is 60.0, then a possible range of uncertainties is +/- 0.5. I think the formulas are supposed to be "statistical" formulas?
    So s_x is an upper bound for the absolute value of the error of x?

    I also don't know how to decipher your base10x expressions. Is this standard notation? What do you mean by it?
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    ^ Yes, I believe that s_x is both the upper and lower bound for the error of x

    The log_{10} a and 10^a a notation is just the form of the value. The base 10 notation isn't really needed, it's just for conversion and ease of use of the formula.

    So, for log_{10} a forms, you would use     s_x=0.434(\frac{s_a}{a})

    And for 10^a a forms, you would use s_x=2.303\cdot x\cdot s_a
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  6. #6
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    Quote Originally Posted by maybealways View Post
    ^ Yes, I believe that s_x is both the upper and lower bound for the error of x

    The log_{10} a and 10^a a notation is just the form of the value. The base 10 notation isn't really needed, it's just for conversion and ease of use of the formula.

    So, for log_{10} a forms, you would use     s_x=0.434(\frac{s_a}{a})

    And for 10^a a forms, you would use s_x=2.303\cdot x\cdot s_a
    Well notation aside, I was able to see where the numbers came from.

    Ignoring sign, write

    s_x = \log_{10}(a+s_a)-\log_{10}(a)

    \dots

    s_x=\log_{10}\left(1+\frac{s_a}{a}\right)

    Use approximation that for small \displaystyle u, 1+u\approx e^u.

    And we can see that 0.434 \approx \log_{10} e.

    Similarly, for the second part,

    s_x = 10^{a+s_a}-10^a

    s_x = 10^a(10^{s_a}-1)

    and for u close to 0, u\approx 1+\log_e u.

    And 2.303 \approx \log_{e} 10.

    The result is not an upper bound but rather an approximation of an upper bound.

    I may not have written/justified as well as possible but that's the general idea.
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  7. #7
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    They've used calculus to convert from f(x+ \Delta x), where \Delta x is the error, to the approximation f(x)+ f '(x)\Delta x using f'(x)= \lim_{\Delta x\to 0}\frac{f(x+ \Delta x)- f(x)}{\Delta x}

    For the natural logarithm, that is particularly easy: If f(x)= ln(x) then f'(x)= 1/x.

    So what they did was convert from "common logarithm", log_{10}(x), to the natural logarithm: [tex]ln(x)= \frac{log_{10}(x)}{log_{10}(e)}[/quote] where e is the base of the natural logarithm: e is approximately 2.718... and ln(e)= 1.

    In any case, they are converting back and forth between base 10 and base "e".

    log_{10}(e)= .434, approximately and
    ln(10)= \frac{1}{log_{10}(e)}= \frac{1}{.4344}= 2.303
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