Exponential Growth and Decay using Logarithms

Exponential Growth and Decay using Logarithms

It has been known how exponential functions can be used to model various growth and decay situations. These included the growth of populations and the decay of radioactive substances. This lesson considers more growth and decay problems, focusing on how logarithms can be used in their solution.

Population Growth

Example 1

The area $A_{t}$ affected by the increase by the insects are given by $A_{t}=200 \times 2^{0.5t}$ m2, where $t$ is the number of days after the initial observation. Find the days taken for the affected area to reach $1000$ m2.

\( \begin{align} \displaystyle
200 \times 2^{0.5t} &= 1000 \\
2^{0.5t} &= \dfrac{1000}{200} \\
&= 5 \\
0.5t &= \log_{2}{5} \\
t &= \dfrac{1}{0.5}\log_{2}{5} \\
t &= 2 \times \dfrac{\log_{10}{5}}{\log_{10}{2}} \\
t &= 4.64 \cdots \\
\end{align} \)
Therefore it takes $5$ days.

Financial Growth

A certain amount of $A_{1}$ is invested at a fixed rate for each compounding period in a financial situation. In this case, the value of the investment after $n$ periods is given by $A_{n+1}=A_{1} \times r^{n}$ where $r$ is the multiple corresponding to the given rate of interest. To find $n$ algebraically, it is required to use $\textit{logarithms}$.

Example 2

$500$ is invested in an account that pays $4.5\%$ per annum, interest compounded monthly. Find how long it takes to reach $\$5000$.

\( \begin{align} \displaystyle
A_{n+1} &= 5000 \\
A_{1} &= 5000 \\
r &= 104.5\% \\
&= 1.045 \\
A_{n+1} &= A_{1} \times r^{n} \\
5000 &= 500 \times 1.045^{n} \\
1.045^{n} &= \dfrac{5000}{500} \\
&= 10 \\
n &= \log_{1.045}{10} \\
&= \dfrac{\log_{10}{10}}{\log_{10}{1.045}} \\
&= 52.311 \cdots \\
\end{align} \)
Therefore it takes $53$ days.


Example 3

The mass $M_{t}$ of radioactive substance remaining after $y$ years given by $M_{t}=6000 \times e^{-0.05t}$ grams. Find the time taken for the mass to halve.

\( \begin{align} \displaystyle
6000 \times e^{-0.05t} &= 3000 \\
e^{-0.05t} &= 3000 \div 6000 \\
&= 0.5 \\
-0.05t &= \log_{e}{0.5} \\
t &= -\dfrac{1}{0.05}\log_{e}{0.5} \\
&= 13.862 \cdots
\end{align} \)
Therefore it takes around $14$ years.

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