Growth and decay problems involve repeated multiplications by a constant number, a common ratio. We can thus use geometric sequences to model these situations.
$$ \large \begin{align} \require{AMSsymbols} \displaystyle
\require{color} \color{red}u_{n} &= u_{1} \times r^{n-1} \\
\require{color} \color{red}u_{n+1} &= u_{1} \times r^{n}
\end{align}$$
Example 1
The initial population of chickens on a farm was $40$. The population increased by $5$% each week.
(a) How many chickens were present after $20$ weeks?
\( \begin{align} \displaystyle
u_{1} &= 40 \\
r &= 1 + 0.05 = 1.05 \\
u_{n} &= 40 \times 1.05^{n-1} \\
u_{20} &= 40 \times 1.05^{20-1} \\
&= 101.078 \cdots
\end{align} \)
There were $101$ chickens.
(b) How long would it take for the population to reach $240$?
\( \begin{align} \displaystyle
40 \times 1.05^{n-1} &= 240 \\
1.05^{n-1} &= 6 \\
n-1 &= \log_{1.05}{6} \\
n &= \log_{1.05}{6} + 1 \\
&= \dfrac{\log_{10}{6}}{\log_{10}{1.05}}\ + 1 \\
&= 37.723 \cdots
\end{align} \)
The population will reach $240$ in the \( 38 \)th week.
Example 2
The population of rabbits on an island at the beginning of $2018$ was $560$. The population has been steadily decreasing by $4$% per year.
(a) Find the population at the beginning of the year \( 2022 \).
\( \begin{align} \displaystyle
u_{1} &= 560 &2018\\
r &= 1-0.04 = 0.96 \\
u_{4} &= 560 \times 0.96^3 &2022 \\
&= 495.45 \cdots
\end{align} \)
The rabbit population will be approximately \( 495 \) at the beginning of \( 2022 \).
(b) In which year would we expect that population to have declined to $140$?
\( \begin{align} \displaystyle
560 \times 0.96^{n-1} &= 140 \\
0.96^{n-1} &= 140 \div 560 \\
&= 0.25 \\
n-1 &= \log_{0.96}{0.25} \\
n &= \log_{0.96}{0.25} + 1 \\
&= \dfrac{\log_{10}{0.25}}{\log_{10}{0.96}} + 1 \\
&= 34.959 \cdots \\
\end{align} \)
The rabbits’ population will be approximately \(1405 \) at \( 35 \) years.
So it will be \( 2053 \).
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