It has been known that how exponential functions can be used to model a variety of growth and decay situations. These included the growth of populations and the decay of radioactive substances. In this lesson we consider more growth and decay problems, focusing particularly on how logarithms can be used in there solution.Exponential Growth and Decay using Logarithms

## Population Growth

### Example 1

The area $A_{t}$ affected by the increase by the insects is given by $A_{t}=200 \times 2^{0.5t}$ m^{2}, where $t$ is the number of days after the initial observation. Find the number of days taken for the affected area to reach 1000 m^{2}. \( \begin{align} \displaystyle
200 \times 2^{0.5t} &= 1000 \\
2^{0.5t} &= \dfrac{1000}{200} \\
&= 5 \\
0.5t &= \log_{2}{5} \\
t &= \dfrac{1}{0.5}\log_{2}{5} \\
t &= 2 \times \dfrac{\log_{10}{5}}{\log_{10}{2}} \\
t &= 4.64 \cdots \\
\end{align} \)

Therefore it takes $5$ days.

## Financial Growth

A certain amount $A_{1}$ 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. In order 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.

## Decay

### 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.