Uncertainty/error question

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Homework Statement



If the relative error of m is 1% and of h is .5%, what is the error of log(mh)?

Homework Equations



d/dx(lnx)=1/x, [tex]\delta[/tex]lnx=[tex]\delta[/tex]x/x

The Attempt at a Solution


[tex]\delta[/tex](ln0.43(mh))=[tex]\delta[/tex](0.43lnm + 0.43lnh)
=[tex]\delta[/tex](0.43ln0.01 + 0.43ln0.005)
=-4.26

i'm pretty sure this isn't right since it's a negative number, i have no idea how to do this
 

Answers and Replies

  • #2
Redbelly98
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I don't see how the 0.43 is coming into this.

First, what's the relative error in m*h?

Second, if you want the error in ln(x), replace x with x0(1±ε). (ε is the relative error in x).

See if you can work it from there.
 

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