Answer
See below
Work Step by Step
The number of bacteria can be described by
$$P(t)=Ce^{kt}$$
with $C,k$ are constant.
Take the logarithm of each side:
$$\ln P(t)=kt+\ln C$$
Let $y=\ln P(t)\\
b=\ln C$
The data points are:
$(0,\ln 3.5);(4,\ln 17);(5,\ln 25.5);(8,\ln 85.5);(10,\ln 189.5)$
Take the logarithm of number of bacteria $P(t)$:
$(0,1.253);(4,2.833);(5,3.239);(8,4.449);(10,5.244)$
Form the matrices:
$A=\begin{bmatrix}
0 & 1\\
4 & 1\\
5& 1\\
8 & 1\\
10 & 1
\end{bmatrix}\\
x=\begin{bmatrix}
k\\b
\end{bmatrix}\\
b=\begin{bmatrix}
1.253\\
2.833\\
3.239\\
4.449\\
5.244
\end{bmatrix}$
Obtain:
$x_0=(A^TA)^{-1}A^Tb\\
=(\begin{bmatrix}
0 & 4 & 5 & 8 & 10\\
1 & 1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
0 & 1\\
4 & 1\\
5& 1\\
8 & 1\\
10 & 1
\end{bmatrix})^{-1}\begin{bmatrix}
0 & 4 & 5 & 8 & 10\\
1 & 1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
1.253\\
2.833\\
3.239\\
4.449\\
5.244
\end{bmatrix}\\
=\begin{bmatrix}
205& 27\\
27 & 5
\end{bmatrix}^{-1}\begin{bmatrix}
0 & 4 & 5 & 8 & 10\\
1 & 1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
1.253\\
2.833\\
3.239\\
4.449\\
5.244
\end{bmatrix}\\
=\frac{1}{296}\begin{bmatrix}
5 & -27\\
-27 & 205
\end{bmatrix}\begin{bmatrix}
0 & 4 & 5 & 8 & 10\\
1 & 1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
1.253\\
2.833\\
3.239\\
4.449\\
5.244
\end{bmatrix}\\
=\frac{1}{296}\begin{bmatrix}
-27 & -7 & -2 & 13 & 23\\
205 & 97 & 70 & -11 & -65
\end{bmatrix}\begin{bmatrix}
1.253\\
2.833\\
3.239\\
4.449\\
5.244
\end{bmatrix}\\
=\frac{1}{296}\begin{bmatrix}
118.309\\
368.597
\end{bmatrix}\\
=\begin{bmatrix}
0.4\\
1.245
\end{bmatrix}\\
k=0.4\\
b=1.245$
Since $b=\ln C\rightarrow C=e^b=e^{1.245}=3.473$
Consequently, $P(t)=Ce^{kt}=3.473e^{0.4t}$
