Calculating Error in Quadrature Equation: y=10^x with dx and dy Values

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To calculate the error in y for the equation y=10^x given an error in x (dx), the correct approach is to differentiate the equation. The derivative yields dy = 10^x * ln(10) * dx, which provides a more accurate error estimate for y. This method resolves the confusion around previous calculations that resulted in unreasonably small errors. The discussion highlights the importance of using proper differentiation techniques in error analysis. Understanding these principles is crucial for applications in cosmological distance calculations.
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I have an equation in the form y=10^x, i have an error in x, dx. I am unsure how to find the associated error in y. dy=10^dx gives to small an error and dy=x-dx doesn't seem logical as a smaller error in x gives a bigger error in y.

(The actual equation is d=10^((m(v)-M(v)+5)/5), i have already have errors for m(v) and M(v) and have combined them by using sqrt(dM(v)^2+dm(v)^2), the equation is used for calculating cosmoligical distances).
 
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Jack_O said:
I have an equation in the form y=10^x, i have an error in x, dx. I am unsure how to find the associated error in y.

Hi Jack_O! :smile:

(try using the X2 tag just above the Reply box :wink:)

Hint: just differentiate … if y = 10x, then dy = (what)dx ? :wink:
 
Hi tiny-tim, doesn't bode well that i had to look up the differential of 10x:frown:

Anyhoo i know get my error in y as 10xln(10)dx, which gives me a much more reasonable answer, thanks for your help:smile:
 
Jack_O said:
Hi tiny-tim, doesn't bode well that i had to look up the differential of 10x:frown:

Hi Jack_O! :smile:

Quick trick: 10x = (eln10)x = exln10 :wink:
 
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