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:
$(0,\ln 100);(1,\ln 24.5);(3,\ln 1.5);(4,\ln 0.4);(6,\ln 0.1)$
Take the logarithm of number of bacteria $P(t)$:
$(0,4.605);(1,3.199);(3,0.405);(4,-0.916);(6,-2.303)$
Form the matrices:
$A=\begin{bmatrix}
0 & 1\\
1 & 1\\
3 & 1\\
4 & 1\\
6 & 1\end{bmatrix}\\
x=\begin{bmatrix}
k\\b
\end{bmatrix}\\
b=\begin{bmatrix}
4.605\\
3.199\\
0.405\\
-0.916\\
-2.303
\end{bmatrix}$
Obtain:
$x_0=(A^TA)^{-1}A^Tb\\
=(\begin{bmatrix}
0 & 1 & 3 & 4 & 6 \\
1 & 1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
0 & 1\\
1 & 1\\
3 & 1\\
4 & 1\\
6 & 1\end{bmatrix})^{-1}\begin{bmatrix}
0 & 1 & 3 & 4 & 6\\
1 & 1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
4.605\\
3.199\\
0.405\\
-0.916\\
-2.303
\end{bmatrix}\\
=\begin{bmatrix}
62& 14\\
14 & 5
\end{bmatrix}^{-1}\begin{bmatrix}
0 & 1 & 3 & 4 & 6 \\
1 & 1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
4.605\\
3.199\\
0.405\\
-0.916\\
-2.303
\end{bmatrix}\\
=\frac{1}{114}\begin{bmatrix}
5 & -14\\
-14 & 62
\end{bmatrix}\begin{bmatrix}
0 & 1 & 3 & 4 & 6 \\
1 & 1 & 1 & 1 & 1
\end{bmatrix}\begin{bmatrix}
4.605\\
3.199\\
0.405\\
-0.916\\
-2.303
\end{bmatrix}\\
=\frac{1}{114}\begin{bmatrix}
-14 & -9 & 1 &6 & 16 \\
62 &48 & 20 & 6 & -22
\end{bmatrix}\\
=\frac{1}{114}\begin{bmatrix}
135.2\\
492.332\end{bmatrix}\\
=\begin{bmatrix}
-1.186\\
4.319
\end{bmatrix}\\
k=-1.186\\
b=4.319$
Since $b=\ln C\rightarrow C=e^b=e^{4.319}=75.113$
Consequently, $P(t)=Ce^{kt}=75.113e^{-1.186t}$
