某种细菌的数量 $n$ 随着时间 $t$（分钟）而增加。记录的 $n$ 和 $t$ 的值如下表所示。

t	2	4	6	8	10
n	3215	3446	3693	3959	4243

已知 $n$ 和 $t$ 由公式 $n = n_{0}\,(2)^{kt}$ 相关联，其中 $n_{0}$ 和 $k$ 是常数。

1. Transform the given equation into a linear form using logarithms.

$n = n_{0}(2)^{kt}$
Taking natural logarithms on both sides:
$\ln n = \ln(n_{0}(2)^{kt})$
$\ln n = \ln n_{0} + \ln(2^{kt})$
$\ln n = (k \ln 2)t + \ln n_{0}$

2. Explain how the linear form corresponds to a straight line graph.

The equation $\ln n = (k \ln 2)t + \ln n_{0}$ is in the form $Y = mX + c$, where $Y = \ln n$, $X = t$, gradient $m = k \ln 2$, and $Y$-intercept $c = \ln n_{0}$.
Thus, plotting $\ln n$ against $t$ will yield a straight line.

3. Calculate the values of the transformed variable $\ln n$ for each given $t$.

4. Draw the straight line graph using the calculated values.

Plot the points $(2, 8.08)$, $(4, 8.14)$, $(6, 8.21)$, $(8, 8.28)$, and $(10, 8.35)$ on the graph paper with $t$ on the horizontal axis and $\ln n$ on the vertical axis, and draw the line of best fit.

1. Identify the relationship between the initial number of bacteria and the graph's vertical intercept

From part (a), the linearised equation is $\ln n = (k \ln 2)t + \ln n_0$. The vertical intercept of the graph of $\ln n$ against $t$ represents $\ln n_0$.

2. Read the intercept from the graph and calculate the initial number of bacteria

From the graph, the vertical intercept is approximately $8.006$.
$\ln n_0 = 8.006$
$n_0 = e^{8.006} \approx 3000$
(Note: If $\lg n$ against $t$ was plotted, the intercept is $\approx 3.477$, giving $n_0 = 10^{3.477} \approx 3000$)

1. Relate the gradient of the straight line graph to the constant $k$

From part (a), the equation is $\ln n = (k \ln 2)t + \ln n_0$.
The gradient of the graph of $\ln n$ against $t$ is $k \ln 2$.

2. Calculate the gradient using points from the table and solve for $k$

Using the points $(2, \ln 3215)$ and $(10, \ln 4243)$:
$\text{Gradient} = \frac{\ln 4243 - \ln 3215}{10 - 2} \approx 0.03467$
$k \ln 2 = 0.03467$
$k = \frac{0.03467}{\ln 2} \approx 0.0500$

1. Set up the equation for a 25% increase in the number of bacteria

An increase of $25\%$ means the number of bacteria becomes $1.25n_0$.
$1.25n_0 = n_0(2)^{kt}$
$1.25 = 2^{kt}$

2. Solve for $t$ using the value of $k$ found in part (c)

Using $k \approx 0.05$ (from part c):
$1.25 = 2^{0.05t}$
Take $\ln$ on both sides:
$\ln 1.25 = 0.05t \ln 2$
$t = \frac{\ln 1.25}{0.05 \ln 2}$
$t \approx 6.44 \text{ minutes}$