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Applications Of Exponentials

1.4 Applications of exponentials (EMBFC)

There are many real world applications that require exponents. For example, exponentials are used to determine population growth and they are also used in finance to calculate different types of interest.

Worked example 16: Applications of exponentials

A type of bacteria has a very high exponential growth rate at \(\text{80}\%\) every hour. If there are \(\text{10}\) bacteria, determine how many there will be in five hours, in one day and in one week?

Exponential formula

\[\text{final population} = \text{initial population} \times (1 + \text{growth percentage})^{\text{time period in hours}}\]

Therefore, in this case: \[\text{final population} = \text{10}\left(\text{1,8}\right)^n\] where \(n =\) number of hours.

In \(\text{5}\) hours

final population = \(\text{10}\left(\text{1,8}\right)^{\text{5}} \approx \text{189}\)

In \(\text{1}\) day = \(\text{24}\) hours

final population = \(\text{10}\left(\text{1,8}\right)^{\text{24}} \approx \text{13 382 588}\)

In \(\text{1}\) week = \(\text{168}\) hours

final population = \(\text{10}\left(\text{1,8}\right)^{\text{168}} \approx \text{7,687} \times \text{10}^{\text{43}}\)

Note this answer is given in scientific notation as it is a very big number.

Worked example 17: Applications of exponentials

A species of extremely rare deep water fish has a very long lifespan and rarely has offspring. If there are a total of \(\text{821}\) of this type of fish and their growth rate is \(\text{2}\%\) each month, how many will there be in half of a year? What will the population be in ten years and in one hundred years?

Exponential formula

\[\text{final population} = \text{initial population} \times (1+\text{growth percentage})^{\text{time period in months}}\]

Therefore, in this case: \[\text{final population} = \text{821}(\text{1,02})^n\] where \(n =\) number of months.

In half a year = \(\text{6}\) months

\(\text{final population} = \text{821}(\text{1,02})^6 \approx \text{925}\)

In \(\text{10}\) years = \(\text{120}\) months

\(\text{final population} = \text{821}(\text{1,02})^{\text{120}} \approx \text{8 838}\)

In \(\text{100}\) years = \(\text{1 200}\) months

\(\text{final population} = \text{821}(\text{1,02})^{\text{1 200}} \approx \text{1,716} \times \text{10}^{\text{13}}\)

Note this answer is also given in scientific notation as it is a very big number.

Applications of exponentials

Exercise 1.7

Nqobani invests \(\text{R}\,\text{5 530}\) into an account which pays out a lump sum at the end of \(\text{6}\) years. If he gets \(\text{R}\,\text{9 622,20}\) at the end of the period, what compound interest rate did the bank offer him? Give answer correct to one decimal place.

\begin{align*} A &= \text{9 622,20} \\ P &= \text{5 530} \\ n &= 6 \end{align*} \begin{align*} A &= P(1 + i)^n \\ \text{9 622,20} &= \text{5 530} (1 + i)^{6}\\ \frac{\text{9 622,20}}{\text{5 530}} &= (1 + i)^{6} \\ \sqrt[6]{\frac{\text{9 622,20}}{\text{5 530}}} &= 1 + i \\ \sqrt[6]{\frac{\text{9 622,20}}{\text{5 530}}} - 1 &= i \\ \therefore i &= \text{0,096709} \ldots \\ &= \text{9,7}\% \end{align*}

The current population of Johannesburg is \(\text{3 885 840}\) and the average rate of population growth in South Africa is \(\text{0,7}\%\) p.a. What can city planners expect the population of Johannesburg to be in \(\text{13}\) years time?

\begin{align*} P &= \text{3 885 840} \\ i &= \text{0,007} \\ n &= 13 \end{align*} \begin{align*} A &= P(1 + i)^n \\ &= \text{3 885 840} (1 + \text{0,007} )^{13} \\ &= \text{4 254 691} \end{align*}

Abiona places \(\text{3}\) books in a stack on her desk. The next day she counts the books in the stack and then adds the same number of books to the top of the stack. After how many days will she have a stack of \(\text{192}\) books?

\(3; 6; 12; 24; 48; \ldots\)

\begin{align*} 3 \times 2^{n-1} &= \text{192} \\ 2^{n-1} &= 64 \\ &= 2^6 \\ \therefore n - 1 &= 6 \\ \therefore n &= 7 \end{align*}

A type of mould has a very high exponential growth rate of \(\text{40}\%\) every hour. If there are initially \(\text{45}\) individual mould cells in the population, determine how many there will be in \(\text{19}\) hours.

\begin{align*} \text{Population } &= \text{Initial population } \times \left( 1 + \text{growth percentage } \right)^{\text{time period in hours} } \\ &= 45 (1 + \text{0,4} )^{19} \\ &= \text{26 893} \end{align*}