# How to calculate uncertainties from a natural exponent graph?

I conducted an experiment in which position of items were shifted on an object, either on the ends of wings of it, or on the base (I'd rather not get too much into what it's about), and the effect on its fall rate over a certain distance was measured. The result was a decay model of the form:

$T(N)=Ae^{-bN}+c$,

where $A=1.44,b=0.132,c=0.303$ and $T =$ Time,$N =$ Number of items added to wings.

However, for each of the times there is an uncertainty of between 0.08 and 0.09 seconds.

So, I asked my teacher for assistance and he explained the following: First you remove the 0.303, and then you can rearrange it as follows:

$T = 1.44*e^{-0.132N}$

$\ln{T} = \ln(1.44*e^{-0.132N})$

$\ln{T} = \ln{1.44} + \ln{e^{-0.132N}}$

$\ln{T} = 0.365 + -0.132N$

$\ln{T} = -0.132N + 0.365$

And thus you have a linear equation.

Then I calculated $\ln{T}$ and $-0.132N + 0.365$ for each value of N, and graphed it in a graphic software, and made error bars of $±((\ln(T+unc)-\ln{(T-unc))/2})$, and thereby can get a best-fit gradient, a maximum possible gradient, and a minimum possible gradient, all in terms of $(ln(T)/(-0.132N + 0.365))$ if I'm not mistaken.

But now for the questions:

1. Why could the (+0.303) simply be removed, and how can that be justified?
2. What do I do with my newly acquired values for the max. and min. gradients?

I'd truly appreciate any help on this!

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