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Logarithmic Uncertainty

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)

\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)

(edited 4 years ago)

Reply 1

Brain Damage

V = 1.9 + 0.1
V_min = 1.8
V_max = 2.0

log(V) = (log(V_max) + log(V_min))/2
Absolute uncertainty = (log(V_max) - log(V_min))/2

log(V) is the average of the max and min, and your uncertainty is half the difference between the max and min.

Reply 2

esrever

Original post by Brain Damage

Makes sense. Thanks!

Reply 3

Eimmanuel

Study Forum Helper

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)

\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}

Reply 4

esrever

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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