Mathematics » Exponents and Surds » Applications Of Exponentials

# Applications of Exponentials

## Applications of Exponentials

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.

## Example

### Question

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: $\text{final population} = \text{10}(\text{1.8})^n$ where $$n =$$ number of hours.

### In $$\text{5}$$ hours

final population = $$\text{10}(\text{1.8})^{\text{5}} \approx \text{189}$$

### In $$\text{1}$$ day = $$\text{24}$$ hours

final population = $$\text{10}(\text{1.8})^{\text{24}} \approx \text{13 382 588}$$

### In $$\text{1}$$ week = $$\text{168}$$ hours

final population = $$\text{10}(\text{1.8})^{\text{168}} \approx \text{7.687} \times \text{10}^{\text{43}}$$

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

## Example

### Question

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.