What Is the Error of Log(mh) When m and h Have Known Relative Errors?

[excel000]

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, $$\delta$$lnx=$$\delta$$x/x

The Attempt at a Solution

$$\delta$$(ln0.43(mh))=$$\delta$$(0.43lnm + 0.43lnh)
=$$\delta$$(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

 

[Redbelly98]

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.

 

[baba89]

I understand that uncertainty and error are important factors to consider when making calculations or measurements. In this case, we are given the relative errors of two quantities, m and h, and are asked to find the error of log(mh).

To find the error of a function, we can use the formula: δf = |f'(x)| * δx, where δf is the error in the function, f'(x) is the derivative of the function, and δx is the error in the input variable.

In this case, the function we are interested in is log(mh). Using the formula above, we can write the error of log(mh) as:

δlog(mh) = |(1/mh) * (δm * h + m * δh)|

Substituting the given values for relative errors, we get:

δlog(mh) = |(1/mh) * (0.01 * h + m * 0.005)|

Since we do not have the values for m and h, we cannot calculate the exact error of log(mh). However, we can make some general observations based on the given information. The relative error of m is larger than that of h (1% vs 0.5%), which means that the error in m will have a greater impact on the overall error of log(mh). Additionally, since log(mh) is a product of m and h, the error will be a combination of the individual errors in m and h.

In conclusion, the error of log(mh) will depend on the values of m and h, but it will likely be larger than the individual errors of m and h due to their combined effect.