Solution to a system of nonlinear equations Do you know any method to solve the following system of nonlinear equations ?
$\begin{equation}
141,3829=A+\frac{B}{323}+5,78C+F323^{E}\\
69,07645=A+\frac{B}{333}+5,81C+F333^{E}\\
40,55085=A+\frac{B}{343}+5,84C+F343^{E}\\
27,92544=A+\frac{B}{353}+5,87C+F353^{E}\\
19,7697=A+\frac{B}{363}+5,89C+F363^{E}
\end{equation}$
where $A,B,C,E,F$ are needs to be determined
 A: Hint
From the first four equations, you can eliminate $a,b,c,f$ (since the equations are linear with respect to these variables) and get them as functions of $e$ (the expressions are not very nice). Replacing their expressions in the fifth equation leads to a single nonlinear equation in $e$ which is monster and needs to be solved.
So, what I should do is to ignore the fifth equation, consider $e$ as a constant, use matrix for solving the first four equation and try to solve the fifth equation. The problem is that there are probaly several solutions.
In fact, doing so does not lead to any solution for $e$ cancelling the fifth equation.
I suppose that this problem is part of a nonlinear regression of data coming from physics or chemistry and the model has been linearized taking logarithms of lhs and rhs. What does it correspond to ? 
It looks like a DIPPR vapor pressure equation $$\log\Big(\frac{1}{P}\Big)=a+\frac{b}{T}+c\log(T)+f T^e$$ which is extremely difficult to fit (except if $e$ is fixed). To me, it looks extremely complex. 
Fixing $e=2$ as often done with DIPPR correlation, I found (this is a linear regression)  $$y=-0.496218 x^2+\frac{8.77839\times 10^7}{x}+372096. \log (x)-2.36971\times 10^6$$ which gives for the lhs $141.203$ , $69.8259$, $39.3843$, $28.7316$, $19.5610$. You must notice that this formulation shows very unusual behavior above $363$.
Taking all of that into account, fix $e$ at a given value, use linear regression to get $a,b,c,f$, compute the sum of squares of residuals and see if this last varies as a function of $e$. If you find an optimum (which must be zero since you have as many data points as parameters), you get the solution.
I shall add that they are things which have been hidden in the post. Obviously, the coefficient of parameter $c$ is the logarithm of inverse of the coefficient of parameter $b$ and probably, the lhs are logarithms of natural numbers.
I shall add a new answer which will take into account of what I percieve to be the reality of this problem.
A: One approach is to use Newton's method for solving $F(x) = 0$.
Given an approximate solution $x_i$, we would ideally like to find $\Delta x$
so that $F(x_i + \Delta x) = 0$.  That's too difficult though, so instead, we use the approximation 
\begin{equation*}
F(x_i + \Delta x) \approx F(x_i) + F'(x_i) \Delta x,
\end{equation*}
and we choose $\Delta x_i$ so that the right hand side is equal to $0$.
Note that $F'(x_i)$ is a matrix.
We then set $x_{i+1} = x_i + \Delta x$.  We then improve $x_{i+1}$, etc.
A: New answer
As I suspected, the post does not reflect the reality of the problem. What is apparently desired in to fit the parameters of the so-called DIPPR vapor pressure equation, namely $$\log(\frac{1}{P})=a+\frac{b}{T}+c\log(T)+f T^e$$ based on only $5$ data points (that is to say as many data points as parameters, forgetting that all measurements are in error). All available data points should have been used for thid regression work. Incidently, the given temperatures have probably be rounded since they probably are 323.15, 333.15 and so on.
As written, in the case where parameter $e$ has a fixed value, the model can be considered as a multilinear model, the independent variables being $\frac{b}{T},\log(T),T^e$  and the dependent variable being $y=-\log(P)$.
So, what can be done is, for many values of $e$, run the multilinear regression from which the sum of squares $(SSQ)$ would be obtained and analyze the behavior of $SSQ(e)$ as a function of $e$. So, for a given $e$, we regress $4$ parameters based on $5$ data points.
If we use $e=0$, the model reduces to $$\log(\frac{1}{P})=a+\frac{b}{T}+c\log(T)$$ for which $SSQ=123,26$.
As soon as an non zero value is assigned to $e$, say $e=0.1$, SSQ drops and we find $SSQ(0.1)=2.29$. But th eproblem is that, increasing $e$ constantly increases $SSQ$.
Now, what must be $e$ ? In most DIPPR vapor pressure correlation $e=2$. Why that ?
To understand, we must go back to history and remember that the  vapor pressure equation results from the integration of the Clausius-Clapeyron equation which write $$\frac{d\Big(log(P)\Big)}{dT}=\int \frac{\Delta H}{T^2}dT$$ and, in the simplest formulations,$$\Delta H=\sum_{i=0}^n a_i T^i$$ and then the classical formulas. DIPPR seleted $e$ as an extra tuning parameter while the polynomial expansions should use $e=1$ .
Nevertheless, using $e=2$, a good fit is obtained but the sum of squares remains of the order of $2.65$ and the corresponding formula is $$y=-2.36971\times 10^6+\frac{8.77839\times 10^7}{T}+372096. \log (T)-0.496218 T^2$$ [($exp=141.38290, calc= 141.20253$),($exp=69.07645, calc= 69.82593$),($exp=40.55085 , calc= 39.38428$),($exp=27.92544   , calc= 28.73161$), ($exp=19.76970   , calc= 19.56099$)].
But we must point out here that, if this model is looking good for interpolation, it does not behave nicely in particulat because of an inflection point at $T=353.758$ and then a too fast decrease even below the highest temperature.
I would suggest to fit the whole set of data points instead of trying to force a five parameter empirical equation to go though five data points which are in errors.
