While solving for a best-fit function with the general form of an exponential equation, that is, $$y=ab^{x-c}+d,$$ I reached these following simultaneous equations as below: \begin{align} ab^{9-c}+d &= 100\\ ab^{14-c}+d &= 150\\ ab^{23-c}+d &= 350\\ ab^{35-c}+d &= 910\\ \end{align}
Which mathematical method can we use to solve for $a, b, c$ and $d$?
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Writing $$ab^{23-c}=50-d$$ and $$ab^{14-c}=910-d$$ dividing each other we get $$b^9=\frac{350-d}{910-d}$$ doing the same with the fourth and the third equation we get $$b^{12}=\frac{910-d}{350-d}$$ eliminating $d$ we get $$b^{12}=\frac{1}{b^9}$$ now you can compute $b$ and so on. Eliminating $d$ as follows: Multiplying by $$350-d$$ we obtain $$350b^{12}-b^{12}d-910=-d$$ so $$350b^{12}-910=db^{12}-d$$
collecting like terms
$$350b^{12}-910=d(b^{12}-1)$$
so9 $$d=\frac{350b^{12}-910}{b^{12}-1}$$
Unfortunately, there is no exact solution to these four equations. Here's why: First, subtract equations (2), (3), (4) by equation (1):
$$ab^{14-c}-ab^{9-c}=50$$ $$ab^{23-c}-ab^{9-c}=250$$ $$ab^{35-c}-ab^{9-c}=810$$
Now, factor out $ab^{9-c}$ from each equation:
$$ab^{9-c}(b^5-1)=50 \ \ (5)$$ $$ab^{9-c}(b^{14}-1)=250 \ \ (6)$$ $$ab^{9-c}(b^{26}-1)=810 \ \ (7)$$
Divide equation (6) by equation (5):
$$\frac{b^{14}-1}{b^5-1}=5\rightarrow b\approx 1.11636$$
Divide equation (7) by equation (5):
$$\frac{b^{26}-1}{b^5-1}=16.2 \rightarrow b\approx 1.0928$$
As you can see, there are two contradictory values of $b$, which is a problem. The best way to deal with this is to just approximate a value of $b$ which is close to both of these values, like saying $b=1.1$. Now, let's plug that back into the original equations.
However, I am going to change the equations a little. Instead of writing $ab^{9-c}$, I will write $mb^9$, where $m=ab^{-c}$. This will help make the equations easier to read, in my opinion:
$$m1.1^9+d=100$$ $$m1.1^{14}+d=150$$ $$m1.1^{23}+d=350$$ $$m1.1^{35}+d=810$$
Now, we have two equations and four unknowns, so there is no direct way to solve for $m$ and $d$. However, we can solve for $m$ and $d$ by using linear regression, where our points are $\{(1.1^9, 100), (1.1^{14}, 150), (1.1^{23}, 350), (1.1^{35}, 810)\}$. From this approach, we get $m \approx 27.1$ and $d \approx 59.6$.
To recap, here's all of the values we have solved for so far:
$$b\approx 1.1$$ $$d\approx 59.6$$ $$m=\frac{a}{b^c}=\frac{a}{1.1^c}\approx 27.1$$
Now, we have definitive, but approximate, values for $b$ and $d$. However, unfortunately, we can not solve the last equation, $\frac{a}{1.1^c}=27.1$ for $a$ and $c$ because there is only one equation and two unknowns. What you do beyond this point is up to you, but I would suggest just making $c=0$ and then solving for $a$ by substituting $c=0$ back into the equation $\frac{a}{1.1^c}=27.1$. I think that is the easiest way to come up for values of $a$ and $c$.
Finally, remember that we did a lot of rounding in this problem, so when you plug $a,b,c,d$ back into the original equations, you will get a lot of rounding error. Again, whether or not this level of rounding error is acceptable in your model is up to you. Good luck!