Evaluate exponential functions with base e

As we saw earlier, the amount earned on an account increases as the compounding frequency increases. The table below shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.

Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies.

Frequency [latex]A\left(t\right)={\left(1+\frac{1}{n}\right)}^{n}[/latex] Value
Annually [latex]{\left(1+\frac{1}{1}\right)}^{1}[/latex] $2
Semiannually [latex]{\left(1+\frac{1}{2}\right)}^{2}[/latex] $2.25
Quarterly [latex]{\left(1+\frac{1}{4}\right)}^{4}[/latex] $2.441406
Monthly [latex]{\left(1+\frac{1}{12}\right)}^{12}[/latex] $2.613035
Daily [latex]{\left(1+\frac{1}{365}\right)}^{365}[/latex] $2.714567
Hourly [latex]{\left(1+\frac{1}{\text{8766}}\right)}^{\text{8766}}[/latex] $2.718127
Once per minute [latex]{\left(1+\frac{1}{\text{525960}}\right)}^{\text{525960}}[/latex] $2.718279
Once per second [latex]{\left(1+\frac{1}{31557600}\right)}^{31557600}[/latex] $2.718282

These values appear to be approaching a limit as n increases without bound. In fact, as n gets larger and larger, the expression [latex]{\left(1+\frac{1}{n}\right)}^{n}[/latex] approaches a number used so frequently in mathematics that it has its own name: the letter [latex]e[/latex]. This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.

A General Note: The Number e

The letter e represents the irrational number

[latex]{\left(1+\frac{1}{n}\right)}^{n},\text{as}n\text{increases without bound}[/latex]

The letter e is used as a base for many real-world exponential models. To work with base e, we use the approximation, [latex]e\approx 2.718282[/latex]. The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties.

Example 7: Using a Calculator to Find Powers of e

Calculate [latex]{e}^{3.14}[/latex]. Round to five decimal places.

Solution

On a calculator, press the button labeled [latex]\left[{e}^{x}\right][/latex]. The window shows [e^(]. Type 3.14 and then close parenthesis, (]). Press [ENTER]. Rounding to 5 decimal places, [latex]{e}^{3.14}\approx 23.10387[/latex]. Caution: Many scientific calculators have an “Exp” button, which is used to enter numbers in scientific notation. It is not used to find powers of e.

Try It 9

Use a calculator to find [latex]{e}^{-0.5}[/latex]. Round to five decimal places.

Solution

Investigating Continuous Growth

So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, e is used as the base for exponential functions. Exponential models that use e as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.

A General Note: The Continuous Growth/Decay Formula

For all real numbers t, and all positive numbers a and r, continuous growth or decay is represented by the formula

[latex]A\left(t\right)=a{e}^{rt}[/latex]

where

  • a is the initial value,
  • r is the continuous growth rate per unit time,
  • and t is the elapsed time.

If > 0, then the formula represents continuous growth. If < 0, then the formula represents continuous decay.

For business applications, the continuous growth formula is called the continuous compounding formula and takes the form

[latex]A\left(t\right)=P{e}^{rt}[/latex]

where

  • P is the principal or the initial invested,
  • r is the growth or interest rate per unit time,
  • and t is the period or term of the investment.

How To: Given the initial value, rate of growth or decay, and time t, solve a continuous growth or decay function.

  1. Use the information in the problem to determine a, the initial value of the function.
  2. Use the information in the problem to determine the growth rate r.
    1. If the problem refers to continuous growth, then > 0.
    2. If the problem refers to continuous decay, then < 0.
  3. Use the information in the problem to determine the time t.
  4. Substitute the given information into the continuous growth formula and solve for A(t).

Example 8: Calculating Continuous Growth

A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year?

Solution

Since the account is growing in value, this is a continuous compounding problem with growth rate = 0.10. The initial investment was $1,000, so = 1000. We use the continuous compounding formula to find the value after = 1 year:

[latex]\begin{cases}A\left(t\right)\hfill & =P{e}^{rt}\hfill & \text{Use the continuous compounding formula}.\hfill \\ \hfill & =1000{\left(e\right)}^{0.1} & \text{Substitute known values for }P, r,\text{ and }t.\hfill \\ \hfill & \approx 1105.17\hfill & \text{Use a calculator to approximate}.\hfill \end{cases}[/latex]

The account is worth $1,105.17 after one year.

Try It 10

A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years?

Solution

Example 9: Calculating Continuous Decay

Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days?

Solution

Since the substance is decaying, the rate, 17.3%, is negative. So, = –0.173. The initial amount of radon-222 was 100 mg, so = 100. We use the continuous decay formula to find the value after = 3 days:

[latex]\begin{cases}A\left(t\right)\hfill & =a{e}^{rt}\hfill & \text{Use the continuous growth formula}.\hfill \\ \hfill & =100{e}^{-0.173\left(3\right)} & \text{Substitute known values for }a, r,\text{ and }t.\hfill \\ \hfill & \approx 59.5115\hfill & \text{Use a calculator to approximate}.\hfill \end{cases}[/latex]

So 59.5115 mg of radon-222 will remain.

Try It 11

Using the data in Example 9, how much radon-222 will remain after one year?

Solution