I have a problem, where the literature says that the following relationship is given:
$$M(z)=a_0z+a_1\text{log}(z)+a_2$$
This is completely fine. When I collect some $(z,M(z))$ pairs, I can do some curve fitting to find the coefficients. The problem that I have is the fact that everyone who graphs the results puts $M(z)$ in the $x$ axis and $z$ in the $y$ axis like shown here:
I know I can just flip the axis to draw the relationship like in Fig. 6.1 , but I want to know what is $z$ in terms of $M(z)$. When I want to invert $M(z)$ in Wolfram, it tells me "No results found in terms of standard mathematical functions"
Hence my question is can I invert this function to put $z$ in terms of $M(z)$?
The graph I get:
$$y(z)=a_0z+a_1\ln(z)+a_2$$ $$\frac{y-a_2}{a_1}=\frac{a_0}{a_1}z+\ln(z)$$ $$e^{\frac{y-a_2}{a_1}}=z\:e^{\frac{a_0}{a_1}z}$$ $$\frac{a_0}{a_1}e^{\frac{y-a_2}{a_1}}=\frac{a_0}{a_1}z\:e^{\frac{a_0}{a_1}z}$$
Let :$\quad\begin{cases} X=\frac{a_0}{a_1}z\\ Y=\frac{a_0}{a_1}e^{\frac{y-a_2}{a_1}} \end{cases}$ $$Y=Xe^X$$ The inverse function of $Y(X)$ is $$X=W(Y)$$ $W$ is the Lambert W function. http://mathworld.wolfram.com/LambertW-Function.html $$\frac{a_0}{a_1}z=W\left(\frac{a_0}{a_1}e^{\frac{y-a_2}{a_1}}\right)$$ $$\boxed{z=\frac{a_1}{a_0}W\left(\frac{a_0}{a_1}e^{\frac{y-a_2}{a_1}}\right)}$$
EXAMPLE : $\quad y=326.3+0.1701x-10.05\ln(x)$