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I would like to express argument $d$ through $f$ and $B$ and to get $d(f,B)$ $$f(d,B)=B\left(\left(\frac{1}{4}-\frac{d}{B}\right)\left( 1-\left(1- e^{-30d/B} \right)^8\right)+\frac{2/B}{3+1/B^2}\left(1- e^{-30d/B} \right)^8\right)$$

How to do it in Mathematica (analytically or numerically)?

ClearAll["Global`*"]
f[d_, B_] := B*((1/4 - d/B)*(1 - (1 - Exp[(-30*d)/B])^8)+
(2/B)/(3 + 1/B^2) (1 - Exp[(-30*d)/B])^8);
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  • $\begingroup$ I'm a bit confused. Is d a function or a variable? Do you want to solve for d and then substitute that result? Did you instead mean f(f(d,B),B)? $\endgroup$
    – alex
    Mar 28, 2023 at 16:15
  • $\begingroup$ @alex, yes d is a variable, I would like to express d through f and B $\endgroup$
    – Mam Mam
    Mar 28, 2023 at 16:39
  • $\begingroup$ Hi @Mam Mam, I'm terribly sorry, but I feel that I do not fully understand your question. I don't think you have added any further addition to what was on your post. $\endgroup$
    – alex
    Mar 28, 2023 at 16:54
  • $\begingroup$ apologies, I just saw that you edited your post since my last comment. Give me a second. $\endgroup$
    – alex
    Mar 28, 2023 at 16:55
  • 1
    $\begingroup$ But the authors of that paper did not claim they can do the inversion analytically. Most probably they did it numerically. $\endgroup$
    – yarchik
    Mar 28, 2023 at 17:25

1 Answer 1

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No chance for mix of exponentials and powers:

Solve[a==f[d,B],d]

This one is working at least with implicit roots

f[d_, B_] := 
 BB* (1 - (1 - Exp[(-30*d)/B])^8) +  (1 - Exp[(-30*d)/B])^8

{{d -> ConditionalExpression[
    1/30 b (2 I \[Pi] ConditionalExpression[1, \[Placeholder]] + 
       Log[Root[-1 + 
           BB + (8 - 8 BB) #1 + (-28 + 28 BB) #1^2 + (56 - 
              56 BB) #1^3 + (-70 + 70 BB) #1^4 + (56 - 
              56 BB) #1^5 + (-28 + 28 BB) #1^6 + (8 - 
              8 BB) #1^7 + (-1 + a) #1^8 &, 1]]), 
    ConditionalExpression[1, \[Placeholder]] \[Element] Integers]

But it is possible to work numerically always with InverseFunction

ClearAll["Global`*"]
f[d_, B_, BB_] := 
 BB* (1 - (1 - Exp[(-30*d)/B])^8) +  (1 - Exp[(-30*d)/B])^8

Plot[Re@InverseFunction[f , 1 , 3][x, 3, 4], {x, 0, 1}]

enter image description here

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