1. ## exponential

13 The volume of chlorine, C litres, in a swimming pool at time t hours after it was placed in
the pool can be modelled by C(t) = C0ekt, t ≥ 0. The volume of chlorine in the pool is
decreasing. Initially the volume of chlorine in the pool was 3 litres, 8 hours later the
volume was 2.5 litres.
a State the value of C0.
b Find the exact value of k.
c Exactly how many litres of chlorine were present in the pool 16 hours after the
chlorine was added to the pool?

i need help on c. how would i do this question.
i sub t=16, Co=3 in the equation but do know what to do next.

2. Originally Posted by rachael
13 The volume of chlorine, C litres, in a swimming pool at time t hours after it was placed in
the pool can be modelled by C(t) = C0ekt, t ≥ 0. The volume of chlorine in the pool is
decreasing. Initially the volume of chlorine in the pool was 3 litres, 8 hours later the
volume was 2.5 litres.
a State the value of C0.
b Find the exact value of k.
c Exactly how many litres of chlorine were present in the pool 16 hours after the
chlorine was added to the pool?

i need help on c. how would i do this question.
i sub t=16, Co=3 in the equation but do know what to do next.
Hello,

to a) $C_0=3\ell$
to b) $c(8)=2.5=3 \cdot e^{k \cdot 8}\ \Rightarrow \ k=\frac{1}{8} \cdot (ln(5)-ln(6))$
that means $k\approx -0.02279...$
to c) Now you've got all the values you need to do the last problem. Plug in t=16 into your equation and you'll get:
$c(16)=3 \cdot e^{\frac{1}{8} \cdot (ln(5)-ln(6)) \cdot 16}=\frac{25}{12} \ell$

Maybe you're a little bit astonished that you get an exakt rational number with c). The problem c) could be done using the results of problem b):
$c(8)=2.5=3 \cdot e^{k \cdot 8} \Longleftrightarrow \frac{5}{6}=e^{k \cdot 8}$
$e^{k \cdot 16}=e^{{k \cdot 8} \cdot 2}=\left(e^{k \cdot 8} \right)^2 = \frac{25}{36}$
so you get: $c(16)=3 \cdot \frac{25}{36}=\frac{25}{12}$
as I've shown above.

Greetings

EB

3. thank you