# Thread: Logarithms and uncertainties: 0.434 and 2.303

1. ## Logarithms and uncertainties: 0.434 and 2.303

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

For: $\displaystyle base 10 x = log_{10} a$

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

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

AND

For: $\displaystyle base 10 x = 10^a a$

$\displaystyle s_x=2.303\cdot x\cdot s_a$

where:
$\displaystyle s_x=$ uncertainty in result
$\displaystyle x=$ result of calculation
$\displaystyle s_a=$ 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?

2. What do you mean by "propagate a few uncertainties?"

3. 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?

4. Originally Posted by maybealways
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 $\displaystyle s_x$ 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?

5. ^ Yes, I believe that $\displaystyle s_x$ is both the upper and lower bound for the error of x

The $\displaystyle log_{10} a$ and $\displaystyle 10^a a$ 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 $\displaystyle log_{10} a$ forms, you would use $\displaystyle s_x=0.434(\frac{s_a}{a})$

And for $\displaystyle 10^a a$ forms, you would use $\displaystyle s_x=2.303\cdot x\cdot s_a$

6. Originally Posted by maybealways
^ Yes, I believe that $\displaystyle s_x$ is both the upper and lower bound for the error of x

The $\displaystyle log_{10} a$ and $\displaystyle 10^a a$ 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 $\displaystyle log_{10} a$ forms, you would use $\displaystyle s_x=0.434(\frac{s_a}{a})$

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

Ignoring sign, write

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

$\displaystyle \dots$

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

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

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

Similarly, for the second part,

$\displaystyle s_x = 10^{a+s_a}-10^a$

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

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

And $\displaystyle 2.303 \approx \log_{e} 10$.

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.

7. They've used calculus to convert from $\displaystyle f(x+ \Delta x)$, where $\displaystyle \Delta x$ is the error, to the approximation $\displaystyle f(x)+ f '(x)\Delta x$ using $\displaystyle f'(x)= \lim_{\Delta x\to 0}\frac{f(x+ \Delta x)- f(x)}{\Delta x}$

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", $\displaystyle log_{10}(x)$, 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".

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

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# ln (e)=2.303

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