# Errror uncertainty for ln(x)

## Homework Statement

What would be the error uncertainty when you take ln of a number. For example ln(10) and the error uncertainty for 10 is ± 1

## The Attempt at a Solution

Is the error uncertainty just (1/10)*2.3? (2.3 is the answer to ln(10))

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rude man
Homework Helper
Gold Member
d/dx ln x = 1/x
what is d(ln x) in terms of dx?
That's for small changes in x.

For larger changes in x, say x → x+a, you can Taylor-series-expand ln (x + a). The first term is of course the above = ln x + a/x , then the higher terms give you better accuracy.

In your case, x = 10 and a = 1, the small-change approximation above gets you to within 0.9980 of the correct answer. Adding just 1 extra term in the series gets you to within 0.9999 of the exact answer. Etc.

NascentOxygen
Staff Emeritus

## Homework Statement

What would be the error uncertainty when you take ln of a number. For example ln(10) and the error uncertainty for 10 is ± 1
So the correct answer lies between ln(9) and ln(11)?

rude man
Homework Helper
Gold Member
So the correct answer lies between ln(9) and ln(11)?
Those are the extremes, yes, but the fractional error distributions differ slightly between x and ln(x).

For small δx/x the fractional change in ln(x) is δ[ln(x)]/ln(x)] ~ [1/ln(x)] δx/x = 0.4343.

E.g. for δx = +1 and x = 10,
fractional change in ln(x) = [ln(11) - ln(10)]/ln(10) = 0.04139 for fractional change in x = 1/10 = 0.1000. So δln(x)/ ln(x) = 0.4139 δx/x.

but if δx = 0.1, fract. change in ln(x) is [ln(10.1) - ln(10)]/ln(10) = 0.004321 for a fractional change in x of 0.1/10 = 0.0100. So δln(x)/ln(x) = 0.4321 δx/x, pretty close to 0.4343.

Which as I said is not a big difference. So with little error, you can say that the fractional error in ln(x) is proportional to the fractional error in x, that ratio being 0.4343, if |δx/x| < 0.1.

But say δx = 5, then
fract. change in ln(x) = [ln(15) - ln(10)]/ln(10) = 0.1761 for δx/x = 0.5, so that ratio is 0.1761/0.5 = 0.3522 which is quite a ways from 0.4343.