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%\title{LogLinearST model}
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\noindent
\begin{large}
\textbf{LogLinearST model} equation is:
\end{large}
\paragraph{}
\noindent
 \begin{eqnarray*}
\begin{array}{l}
\displaystyle \log_{10}(Nt) = log_{10}(A \times B  \times C  + 10^{log_{10}(N_r)})
\end{array}
\end{eqnarray*}
\noindent
where
\begin{eqnarray*}
\begin{array}{l}
\displaystyle A= 10^{log_{10}(N_0)} - 10^{log_{10}(N_r)} \\[3mm]
      \displaystyle B = exp( -k_{max} \times t) \\[3mm]
     \displaystyle C = D / (1 +(D-1) \times B) \\[3mm]
\end{array}
\end{eqnarray*}
with
\begin{eqnarray*}
\begin{array}{l}
\displaystyle D = exp(k_{max} \times S1) \\[3mm]
\end{array}
\end{eqnarray*}
\noindent
 $t$ is time, $log_{10}$ is base 10 logarithm. The parameters  to estimate are $k_{max}$, $S1$, $log_{10}(N_0)$ and $log_{10}(N_r)$.
\newline
The noisy output is defined as:
\begin{displaymath}
\log_{10}(Nt)=\mathcal{N}(\log_{10}(Nt), \%noise)
\end{displaymath}
i.e random number from the normal distribution with mean parameter $\log_{10}(Nt)$ and standard deviation parameter $\%noise$.

\paragraph{}
\begin{flushleft}
\textbf{Example of LogLinearST curve}\\
Time unit is mn. Maximal time is 60mn.
$k_{max}=0.5$, $S1=8$, $log_{10}(N_0)=6$ and $log_{10}(N_r)=0.5$.
\end{flushleft}

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