a formula with natural logarithm in tcl - tcl

Question

This formula will be coded in tcl.
Y= Intercept + Slope1*(X – X0) + (Slope2 – Slope1)*Delta*ln(1+exp((X-X0)/Delta))
I searched that ln() in this formula can be coded using log() in tcl.
However, I cannot get the correct output in tcl.
Here is what I coded, could you please take a look? Any comments? Thank you.
set Intercept 0.7416
set Slope1 52.42
set x0 0.01491
set Slope2 0.2533
set Delta 0.002275
set y_frac [expr {
$Intercept + $Slope1*($x/$tend - $x0/$tend)
+ ($Slope2-$Slope1) * $Delta * log10(1.0 + exp(($x/$tend - $x0/$tend)/$Delta))
}]
set y [expr {$y_frac *100.0}] ​

Answer 1

The log() function computes the natural logarithm of its argument (just as the exp() function is its inverse, computing ex). That means your expression becomes (with the simplest of transformations) just:
set Y [expr {
$Intercept + $Slope1*($X – $X0) + ($Slope2 – $Slope1)*$Delta*log(1+exp(($X-$X0)/$Delta))
}]
I'd refactor the expression a little, as deltas are natural things to extract:
set dX [expr { $X - $X0 }]
set dSlope [expr { $Slope2 - $Slope1 }]
set Y [expr { $Intercept + $Slope1*$dX + $dSlope*$Delta*log(1+exp($dX/$Delta)) }]
At this point, it's probably easier to define a procedure to encapsulate this.
proc computeY {y} {
global Intercept Slope1 Slope2 Delta x0
set dX [expr { $x - $x0 }]
set dSlope [expr { $Slope2 - $Slope1 }]
set Y [expr { $Intercept + $Slope1*$dX + $dSlope*$Delta*log(1+exp($dX/$Delta)) }]
return [expr { $Y * 100.0 }]; # Your final rescaling
}