Given the non-linear equation y = A + B e C t and 3 sets of points: ( y 1 , t 1 ) , ( y 2 , t 2 ) , ( y 3 , t 3 ) ,, can the variables A, B, and C be calculated analytically? y 1 = A + B e C t 1 y 2 = A + B e C t 2 y 3 = A + B e C t 3 y 1 − y 2 = B e C t 1 − B e C t 2 y 2 − y 3 = B e C t 2 − B e C t 3 y 1 − y 2 y 2 − y 3 = e C t 1 − e C t 2 e C t 2 − e C t 3 This is where I am stuck. Without using approximations, I can't reduce this function to determine the variable C. Is this possible?

Question

Given the non-linear equation   y  =  A  +  B        e          C      t       and   3 sets of points:   (      y    1    ,      t    1    )  ,  (      y    2    ,      t    2    )  ,  (      y    3    ,      t    3    )  ,, can the variables   A,   B, and   C be calculated analytically?                              y          1                =        A        +        B                          e                      C                          t              1                                                                    y          2                =        A        +        B                          e                      C                          t              2                                                                    y          3                =        A        +        B                          e                      C                          t              3                                                                        y          1                −                  y          2                =        B                          e                      C                          t              1                                      −        B                          e                      C                          t              2                                                                    y          2                −                  y          3                =        B                          e                      C                          t              2                                      −        B                          e                      C                          t              3                                                                                                  y              1                        −                          y              2                                                          y              2                        −                          y              3                                      =                                                        e                              C                                  t                  1                                                      −                                      e                              C                                  t                  2                                                                                                    e                              C                                  t                  2                                                      −                                      e                              C                                  t                  3                                                                        This is where I am stuck. Without using approximations, I can&#039;t reduce this function to determine the variable   C. Is this possible?

xxsailojaixxv5 · Accepted Answer

In the most general case, as written, you cannot extract   C from the third equation since it is highly transcendental. The only solution would be a numerical method such as Newton which would work without any problem provided a reasonable guess; a plot of the function (the third equation) as a function of   C will allow you to locate the root and start iterating from this value.However, there are cases where this is feasible. The simplest I found corresponds to       t    2    =  2      t    1   and      t    3    =  3      t    1  . For such a case, simplifying the rhs, you should arrive to  C  =            log      ⁡              (                                                      y              2                        −                                    y              3                                                                    y              1                        −                                    y              2                                      )                      t      1      Similarly, if       t    2    =  2      t    1   and       t    3    =  4      t    1  , you should arrive to  C  =            log      ⁡              (                              −                                                      y                1                            −                                          y                2                                                                                y                1                            +              3                                          y                2                            −              4                                          y                3                                      −                                    y              1                        +                                    y              2                                            2            (                                    y              1                        −                                    y              2                        )                          )                      t      1

Given the non-linear equation y = A + B <mi mathvariant="normal">e C

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