# How to determine the parameters of a linearized model when the line has negative slope and intercept?

I have the following data:

Sample $$t$$ $$q_t$$ $$\ln\left(q_e-q_t\right)$$
A 0 0.00 0.31
A 60 0.97 -0.92
A 90 1.30 -2.71
A 120 1.33 -3.40
A 180 1.28 -2.48
A 240 1.35 -4.09
A $$q_e=$$ 1.37
B 0 0.00 0.64
B 60 1.12 -0.24
B 90 1.77 -2.01
B 120 1.83 -2.71
B 180 1.88 -4.09
B 240 1.80 -2.30
B $$q_e=$$ 1.90

The data, from both sample A and sample B, were adjusted to the following model in its linearized form, to determine the value of the parameters $$k$$ and $$q_e$$: $$\ln\left(q_e-q_t\right)=\ln q_e-kt \rightarrow \rm linearized\ form$$ $$q_t=q_e\left[1-\exp\left(-kt\right)\right]\rightarrow \rm original\ model$$ I did the regressions and these are the results: According to my mathematical knowledge, a straight line is defined by the equation $$y = b+mx$$. With that the calculation of the parameters $$k$$ and $$q_e$$ would be:

Straight line $$y$$ $$=$$ $$b$$ $$+$$ $$m$$ $$x$$
Model $$\ln\left(q_e-q_t\right)$$ $$=$$ $$\ln q_e$$ $$-$$ $$k$$ $$t$$
• $$q_e=e^{\rm intercept}=\exp(\rm intercept)$$
• $$k=-\rm slope$$

Following this procedure, I estimated the values of $$k$$ and $$q_e$$ for each sample and then I plotted it in a non-linear form:

Sample $$k$$ $$q_e$$ Plotted function
A 0.02 0.73 $$q_t=0.73\left[1-\exp(-0.02t)\right]$$
B 0.02 1.02 $$q_t=1.02\left[1-\exp(-0.02t)\right]$$ Looking at the graph, I believe that I made a mistake in one of my procedures and I would like your help, because I believe that the non-linear curves should be closer to the points.

It is difficult to reproduce your calculus and spot the "mistake" because your calculus is not made with sufficient accurency and the numbers in the tables are too roughtly rounded.

For example, sample A , t=240 ; qt=1.35 ; qe=1.35 ; it is written ln(qe-qt)=-4.09

The correct value is ln(qe-qt)=-3.912023

This is a big deviation for the further calculus. Remember : It is allowed to round the final result of the calculus. But it is not allowed to round all along the intermediate steps of the calculus.

Nevertheless, it seems that the main cause of trouble in your method is the initial guessing of qe=1.35

At the end of calculus you got qe=0.73 ; This is very far from 1.35 ; Morover qe=0.73 is not compatible with the calculus of ln(qe-qt) which is then not real.

A better accurate calculus is shown below, starting with qe=1.35053

The steps of the numerical calculus are given with a sufficient accuracy to allow everybody to check. The above initial guess for qe is not quite satisfying but better than before. That is why the fitting of the curves appears much better.

In fact your method involves an approach by trial and error to adjust the value of qe. This is very tiresome. Generally one use some non-linear method of regression.