If i have a=exp x, and i know that my uncertainty in a is 3% ( at all

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To calculate the uncertainty in x when a = exp(x) and the uncertainty in a is 3%, one should use error propagation. The relationship can be expressed as x = ln(a), allowing for the derivative to be easily found. The absolute uncertainty in x can be derived from the percentage uncertainty in a, leading to the conclusion that dx = da/a. However, it is important to note that the units must be consistent, as x cannot have units of mm if a = 2 does not correspond to x = 3. Thus, the absolute uncertainty in x is indeed equal to the percentage uncertainty in a.
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if i have a=exp x, and i know that my uncertainty in a is 3% ( at all points). At a point where a=2, how do we calculate the uncdrtainty in x.

is the absolute value = (3% of ln 2)/ln 2, which equals 0.03mm, which equals 1% (if x is 3mm)

thanks
 
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x cannot have units of mm in that equation, and a=2 does not correspond to x=3.

Use error propagation - write the expression as x=ln(a), the derivative with respect to a is easy to find then.
The formula looks good, but the result has the problem mentioned above.
 
so dx = da/a?

so the absolute uncertainty in x is equal to the percentage uncetainty in a?
 
So I know that electrons are fundamental, there's no 'material' that makes them up, it's like talking about a colour itself rather than a car or a flower. Now protons and neutrons and quarks and whatever other stuff is there fundamentally, I want someone to kind of teach me these, I have a lot of questions that books might not give the answer in the way I understand. Thanks
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