Logarithmic Uncertainty Watch

esrever
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Suppose V = 1.9 \pm 0.1. How to find absolute uncertainty for \log_{10}(V)?

I know two methods which are as follows:

1) \log_{10}(1.9 + 0.1) - \log_{10}(1.9)

2) \dfrac{0.1}{1.9\ln(10)}

Both methods produce approximately the same answers. Which one do I use? Or does it not matter? (For CIE exam board)
Last edited by esrever; 1 month ago
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Brain Damage
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V = 1.9 + 0.1
Vmin = 1.8
Vmax = 2.0

log(V) = (log(Vmax) + log(Vmin))/2
Absolute uncertainty = (log(Vmax) - log(Vmin))/2

log(V) is the average of the max and min, and your uncertainty is half the difference between the max and min.
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esrever
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(Original post by Brain Damage)
V = 1.9 + 0.1
Vmin = 1.8
Vmax = 2.0

log(V) = (log(Vmax) + log(Vmin))/2
Absolute uncertainty = (log(Vmax) - log(Vmin))/2

log(V) is the average of the max and min, and your uncertainty is half the difference between the max and min.
Makes sense. Thanks!
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Eimmanuel
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(Original post by esrever)
Suppose V = 1.9 \pm 0.1. How to find absolute uncertainty for \log_{10}(V)?

I know two methods which are as follows:

1) \log_{10}(1.9 + 0.1) - \log_{10}(1.9)

2) \dfrac{0.1}{1.9\ln(10)}

Both methods produce approximately the same answers. Which one do I use? Or does it not matter? (For CIE exam board)
It does not matter.

But your method 1 should be replaced by Brain Damage method:

 \dfrac{\log_{10}(1.9 + 0.1) - \log_{10}(1.9 - 0.1)}{2}
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esrever
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(Original post by Eimmanuel)
It does not matter.

But your method 1 should be replaced by Brain Damage method:

 \dfrac{\log_{10}(1.9 + 0.1) - \log_{10}(1.9 - 0.1)}{2}
Thanks!
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