Topic 16 Exponential Functions

16.1 Half-life

Half-life is the time required for a quantity to reduce to half of its initial value.

A certain pesticide is used against insects. The half-life of the pesticide is about 12 days. After a month how much would left if the initial amount of the pesticide is 10 g? Can you write a function for the remaining quantity of the pesticide after \(t\) days?

More examples on exponential functions can be found on

16.2 Definition and Graphs of Exponential Functions

Let \(b\) be a positive number other than \(1\) (i.e. \(b>0\) and \(b\neq 1\)). The exponential function \(f\) of \(x\) with the base \(b\) is defined as \[ f(x)=b^x\quad\quad\text{or}\quad\quad y=b^x. \]

Graphs of exponential functions:

The exponential function \(f(x)=b^x\) is a one-to-one function: any vertical line or any horizontal line crosses the graph at most once. Equivalently, the equation \(b^x=c\) has at most one solution for any real number \(c\).

16.3 The Natural Number \(e\)

The natural number \(e\) is the number to which the quantity \(\left(1+\dfrac1n\right)^n\) approaches as \(n\) takes on increasingly large values. Approximately, \(e\approx2.718281827\).

16.4 Compound Interests

After \(t\) years, the balance \(A\) in an account with a principal \(P\) and annual interest rate \(r\) is given by the following formulas:

  1. For \(n\) compounding periods per year: \(A=P\left(1+\dfrac{r}{n}\right)^{nt}\).
  2. For compounding continuously: \(A=Pe^{rt}\).

Example 16.1 A sum of \(\$10,000\) is invested at an annual rate of \(8\%\), Find the balance, to the nearest hundredth dollar, in the account after \(5\) years if the interest is compounded

  1. monthly,
  2. quarterly,
  3. semiannually,
  4. continuous.


  1. Find values of \(P\), \(r\), \(t\) and \(n\). In this case, \(P=10,000\), \(r=8\%=0.08\), \(t=5\) and \(n\) depends compounding.
  2. Plug the values in the formula and calculate.
  3. ``Monthly’’ means \(n=12\). Then \[ A=10000\left(1+\frac{0.08}{12}\right)^{5\cdot 12}\approx 14898.46. \]
  4. ``Quarterly’’ means \(n=4\). Then \[ A=10000\left(1+\frac{0.08}{4}\right)^{5\cdot 4}\approx 14859.47. \]
  5. ``semiannually’’ means \(n=2\). Then \[ A=10000\left(1+\frac{0.08}{2}\right)^{5\cdot 2}\approx 14802.44. \]
  6. For continuously compounded interest, we have \[ A=10000e^{0.08\cdot 5}\approx 14918.25. \]

In the compounded investment module, the \(\frac rn\) is an approximation of the period interest rate. Indeed, if the period rate \(p\) satisfies the equation \((1+p)^n=1+r\), or equivalently \(p=\sqrt[n]{1+r} - 1\). Using the formula \((1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\cdots +x^n\), one may approximately replace \(1+r\) by \((1+\frac rn)\) and obtain the approximation \(p\approx \frac rn\).

Example 16.2 The population of a country was about 0.78 billion in the year 2015, with an annual growth rate of about 0.4%. The predicted population is \(P(t)=0.78(1.004)^t\) billions after \(t\) years since 2015. To the nearest thousandth of a billion, what will the predicted population of the country be in 2030?


The population is approximately \[ P(15)=0.78(1.004)^{15}\approx 0.828 \quad \text{billions}. \]

16.5 Practice

Problem 16.1 The value of a car is depreciating according to the formula: \(V=25000(3.2)^{-0.05x}\), where \(x\) is the age of the car in years. Find the value of the car, to the nearest dollar, when it is five years old.

Problem 16.2 A sum of $20,000 is invested at an annual rate of 5.5%, Find the balance, to the nearest dollar, in the account after 5 years subject to

  1. monthly compounding,
  2. continuously compounding.

Problem 16.3 Sketch the graph of the function and find its range.

  1. \(f(x)=3^x\)
  2. \(f(x)=\left(\frac13\right)^x\)

Problem 16.4 Use the given function to compare the values of \(f(-1.05)\), \(f(0)\) and \(f(2.4)\) and determine which value is the largest and which value is the smallest. Explain your answer.

  1. \(f(x)=\left(\frac{5}{2}\right)^x\)
  2. \(f(x)=\left(\frac23\right)^x\)