How to find $y(x,z)$ from the given set of data? I have the following set of data:



$x_1,x_2,x_3,\dots,x_m$ are increasing in arithmetical  progression.

$y_1,y_2,y_3,\dots,y_n$ are increasing in arithmetical  progression.

$z(x_i,y_1),z(x_i,y_2),z(x_i,y_3),\dots,z(x_i,y_n)$ are exponentially increasing for $i=1,2,3,\dots,m$.

$z(x_1,y_i),z(x_2,y_i),z(x_3,y_i),\dots,z(x_m,y_i)$ are exponentially decreasing for $i=1,2,3,\dots,n$.

All values in the table are positive.

The required is to find the value of $y$ in terms of $x$ and $z$, [i.e. $y(x,z)$].


  How to find $y(x,z)$ from the given set of data?




Consider the following example:

The data set is:



Say we need to find the value of $y$ when $x=172$ and $z=3.1527$. Clearly, from the table, we can find that $y(172,3.1527)=25.50$.

What if we want to find $y(150,3.1729)$? which is not in the table

The answer is $y(x,z)=x(-1+\ln(z))$, so $y(150,3.1729)=150(-1+\ln(3.1729))=23.1969$

I actually know the expression $\boxed{y(x,z)=x(-1+\ln(z))}$ because I made this example to demonstrate my problem.

What if we do not know that relation, how should we obtain it?



This problem is related to my research in the chemical laboratory, where the values of $x$'s are the speeds of the centrifuge (rpm), the values of $y$'s are the volumes of the samples (mL), and the values of $z$'s are the length in (mm) of the extracted liquid in the pictures that are taken by a camera.



Regardless of my research, and regardless of the units (rpm, mL, mm), can we generalise a method?



Any help would be really appreciated. THANKS!
HINT.

Formally, is described the model
$$z(x,y) = e^{a+bx+cy},$$
or
$$\ln z = a+bx+cy,\tag1$$
where $b&lt;0, c&gt;0,$
$$x_i=12+2.25i,\quad y_i= 100 + 24j.$$

The table data model is
$$z_{i,j} = e^{\large \frac xy+1}\tag2$$



Assuming the discrepancy in the form of
$$d(a,b,c) = \sum w_{i,j}(\ln z_{i,j} - a - bx_i - Cy_j)^2,\tag3$$
where $w_{ij}$ is the arbitrary matrix of weights,

one can get the point (A,B,C), which provides $\min d(a,b,c)$ in accordance with the described model.

This point is the stationary point of $d(a,b,c).$ 

So $\operatorname{grad} d(A,B,C) = 0,$ or
\begin{cases}
\sum w_{i,j}\ (\ln z_{i,j} - A - Bx_i - Cy_j)=0\\
\sum w_{i,j}\ i\ (\ln z_{i,j} - A - Bx_i - Cy_j)=0\\
\sum w_{i,j}\ j\ (\ln z_{i,j}  - A - Bx_i - Cy_j) = 0.\tag4
\end{cases}

This leads to the linear system
\begin{cases} 
S_{00}A + S_{10} B + S_{01}C = R_{00}\\
S_{10}A + S_{20} B + S_{11}C = R_{10}\\
S_{01}A + S_{11} B + S_{02}C = R_{01},\tag5
\end{cases}

where 
$$S_{kl} = \sum w_{ij}x_i^k y_j^l,\quad R_{kl} = \sum w_{ij} x_i^k j^l \ln z_{ij}.\tag6$$

Using weight array $w=1$ gives



and looks unusable. This situation is happened, because the data table does not correspond to the model.

However, applying of the weight array in the form of
$$w_{ij}=e^{-\left(\Large\frac{5(x_i-150)}{24}\right)^2-\left(\Large\frac{2(y_j-23.25)}{2.25}\right)^2}$$

localizes the model and gives



So the estimation is
$$Y = \dfrac{\ln z -A-Bx}C 
\approx \dfrac{\ln z - 1.13517\ 52307 + 0.00091\ 33448x}{0.00675\ 67568}, \tag7$$

and the result



looks suitable.

Note that the constants "2" and "5" in the $w$ formula were obtained empirically,
with the goal of the good approximation of the table data near of the expected point. If the table data model corresponds with the given model better, then these constants can be decreased or $w=1$ can be used.
If $z$ follows geometric progressions in terms of $x,y$, then $\log z$ follows arithmetic progressions.

If you keep one of the coordinates fixed, 

$$y=\text{Cst}\to\log z=ax+b$$

$$x=\text{Cst}\to\log z=cy+d.$$

Combining the two, we have a bivariate linear relation

$$\log z=px+qy+r.$$

The parameters $p,q,r$ can be determined from three data points, solving a $3\times3$ system.

Obviously,

$$y=\frac{\log z-px-r}q.$$