If you want to model the population with the formula:
$$
P(t)=P_0 e^{kt},
$$
where $P_0$ is the initial population and $k$ is the exponential growth constant,
then you must first find the values of $P_0$ and $k$.
We are told $P_0=1$, so
$$
P(t)= e^{kt}
$$
Let's now solve for $k$:
We know that after 5 minutes, the population is $2 $, so
$$
2 = e^{5k}
$$
Solving the above for $k$ gives $k={\ln 2\over 5}$.
So
$$
P(t)=1\cdot e^{ {\ln 2\over 5}t}
$$
After 96 minutes, the "population" is
$$
P(96)=1\cdot e^{ {\ln 2\over 5}96}= e^{19.2 {\ln 2 } }\approx 602,248.763.
$$
But, as @Henning Makholm points out in the comment below, this isn't realistic. The population at 96 minutes, assuming a bacteria splits in 2 every 5 minutes, is the population at $95$ minutes:
$P(95)= e^{ {\ln 2\over 5}95}= e^{\ln 2\cdot19}=2^{19}$.
This could have been obtained more simply: 96 minutes is 19 doubling periods (plus an extra minute where no new bacteria are formed).
$\ \ \ $after 5 minutes, one doubling period, the population is 2.
$\ \ \ $after another 5 minutes, one more doubling period, the population is 4.
$\ \ \ $after another 5 minutes, one more doubling period, the population is 8.
$\ \ \ \ \ \ \ \ \vdots$
At 95 minutes, 19 doubling periods,
the population will be $2^{19}$. At 96 minutes, the population will be $2^{19}$.
Let me take the second problem you gave in your last comment.
First of all, we shall note $T(t)=T_0~a^t$ and $S(t)=S_0~b^t$ the populations of termites and spiders as a function of time.
Concerning the termites, the first informations you give translate to $$T(0)=100$$ $$T(4)=200$$ From these, $T_0=100$ and $a=\sqrt[4]{2}~~ (\simeq 1.18921)$.
Concerning the ratio between termites and spiders, the next informations translate to $$T(3)=2~S(3)$$ $$T(8)=4~S(8)$$ This write $$T_0~ a^3=2~ S_0~ b^3$$ $$T_0~ a^8=4~ S_0 ~b^8$$ and taking the ratio $$\frac{T_0~ a^8}{T_0~ a^3}=\frac{4~S_0~ b^8}{2~S_0~ b^3}$$ which simplifies to $a^5=2~b^5$ or $2^{\frac{5}{4}}=2~b^5$ or $2^{\frac{1}{4}}=b^5$ and finally $b=\sqrt[20]{2} ~~(\simeq 1.03526)$.
So now, the last question translates to $$S(t)=3~S(0)$$ As you noticed, we do not need to care about $S_0$ which eleminates from the equation. So what we need is to find $t$ such that $$\frac {S(t)} {S(0)}=\frac {S_0~b^t}{S_0~b^0}=b^t=3$$ Taking logarithms of both sides and replacing $b$ by its value, we arrive to $$t=\frac{20 \log (3)}{\log (2)} \simeq 31.6993$$
Best Answer
Hint: In general the equation is $y=y_0\cdot a^t$, where $y$ is the poputlation of the bacteria after $t$ hours and $y_0=60$ is the initial population. Now you can deduce an equation from the following information:"A certain type of bacteria doubles every $6.5$ hours"
$$120=60\cdot a^{6.5}$$
Solve the equation for $a$ and obtain the function of the population.