5.2: Exponential Functions
- Page ID
- 89582
- Evaluate exponential functions.
- Use compound interest formulas.
- Evaluate exponential functions with base \(e\).
India is the second most populous country in the world with a population of about \(1.25\) billion people in 2013. The population is growing at a rate of about \(.2\%\) each year. If this rate continues, the population of India will exceed China’s population by the year 2031. When populations grow rapidly, we often say that the growth is “exponential,” meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions, which model this kind of rapid growth.
Exponential Growth and Decay
What exactly does it mean to grow exponentially? For us to gain a clear understanding of exponential growth, let us compare exponential growth with linear growth. We will construct two functions. The first function, \(f(x)=2^x\), is exponential. The second function, \(g(x)=2x\), is linear. We will start with an input of \(0\), and increase each input by \(1\). The corresponding outputs are shown in Table \(\PageIndex{1}\). From Table \(\PageIndex{1}\), we can see that for the exponential function, \(f(x)\) that as we increase the input by 1 that the output doubles. The linear function, \(g(x)\), is a constant increase by 2. From Table \(\PageIndex{1}\) we can conclude that exponential growth dwarfs linear growth.
- Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain.
- Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain.
| \(x\) | \(f(x)=2^x\) | \(g(x)=2x\) |
|---|---|---|
| 0 | 1 | 0 |
| 1 | 2 | 2 |
| 2 | 4 | 4 |
| 3 | 8 | 6 |
| 4 | 16 | 8 |
| 5 | 32 | 10 |
| 6 | 64 | 12 |
For exponential growth, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding \(2\) to the output whenever the input was increased by one.
The general form of an exponential function is \(f(x)=ab^x\), where \(a\) is any nonzero number, \(b\) is a positive real number not equal to \(1\).
- If \(b>1\),the function grows at a rate proportional to its size. This is called exponential growth.
- If \(0<b<1\), the function decays at a rate proportional to its size. This is called exponential decay.
Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time.
Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time .
Let’s look at the function \(f(x)=2^x\) from our example. We will create a table (Table \(\PageIndex{2}\)) to determine the corresponding outputs over an interval in the domain from \(−3\) to \(3\).
| \(x\) | \(−3\) | \(−2\) | \(−1\) | \(0\) | \(1\) | \(2\) | \(3\) |
|---|---|---|---|---|---|---|---|
| \(f(x)=2^x\) | \(2^{−3}=\dfrac{1}{8}\) | \(2^{−2}=\dfrac{1}{4}\) | \(2^{−1}=\dfrac{1}{2}\) | \(2^0=1\) | \(2^1=2\) | \(2^2=4\) | \(2^3=8\) |
Let us examine the graph of \(f\) by plotting the ordered pairs from Table \(\PageIndex{2}\) and then make a few observations in Figure \(\PageIndex{1}\).
Let’s define the behavior of the graph of the exponential function \(f(x)=2^x\) and highlight some its key characteristics.
- the domain is \((−\infty,\infty)\),
- the range is \((0,\infty)\),
- as \(x\rightarrow \infty\), \(f(x)\rightarrow \infty\),
- as \(x\rightarrow −\infty\), \(f(x)\rightarrow 0\),
- \(f(x)\) is always increasing,
- the graph of \(f(x)\) will never touch the x-axis because base two raised to any exponent never has the result of zero.
- \(y=0\) is the horizontal asymptote.
- the y-intercept is \((0,1)\).
For any real number \(x\), an exponential function is a function with the form
\[f(x)=ab^x \nonumber \]
where
- \(a\) is a non-zero real number called the initial value and
- \(b\) is any positive real number such that \(b≠1\).
- The domain of \(f\) is all real numbers.
- The range of \(f\) is all positive real numbers if \(a>0\).
- The range of \(f\) is all negative real numbers if \(a<0\).
- The y-intercept is \((0,a)\),and the horizontal asymptote is \(y=0\).
\(b\) is called the base of the function and \(x\) is called the exponent.
Note that an exponential function does not intersect its horizontal asymptote. This is because there is no value for \(x\) that we can input into the function that would yield an output of 0, \(ab^x \neq 0\).
Which of the following equations are not exponential functions?
- \(f(x)=4^{3(x−2)}\)
- \(g(x)=x^3\)
- \(h(x)=\left(\dfrac{1}{3}\right)^x\)
- \(j(x)=(−2)^x\)
Solution
By definition, an exponential function has a constant base and a variable exponent. Thus, \(g(x)=x^3\) does not represent an exponential function because the base is variable and the exponent is constant.
Recall that the base \(b\) of an exponential function is always a positive constant, and \(b≠1\). Thus, \(j(x)={(−2)}^x\) does not represent an exponential function because the base, \(−2\), is less than \(0\).
\(f(x)\) and \(h(x)\) are exponential functions.
Which of the following equations represent exponential functions?
- \(f(x)=2x^2−3x+1\)
- \(g(x)={0.875}^x\)
- \(h(x)=1.75x+2\)
- \(j(x)={1095.6}^{−2x}\)
- Answer
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\(g(x)={0.875}^x\) and \(j(x)={1095.6}^{−2x}\) represent exponential functions. The others are not exponential functions.
Recall that the base of an exponential function must be a positive real number other than \(1\). Why do we limit the base, b ,to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:
Let \(b=−9\) and \(x=\dfrac{1}{2}\). Then \(f(x)=f\left(\dfrac{1}{2}\right)={(−9)}^{\dfrac{1}{2}}=\sqrt{−9}\), which is not a real number.
Why do we limit the base to positive values other than \(1\)? Because base \(1\) results in the constant function. Observe what happens if the base is \(1\):
Let \(b=1\). Then \(f(x)=1^x=1\) for any value of \(x\).
Evaluating Exponential Functions
To evaluate an exponential function with the form \(f(x)=b^x\), we simply substitute \(x\) with the given value, and calculate the resulting power. For example:
Let \(f(x)=2^x\). What is \(f(3)\)?
\[\begin{align*} f(x)&= 2^x\\ f(3)&= 2^3 \qquad \text{Substitute } x=3\\ &= 8 \qquad \text{Evaluate the power} \end{align*}\]
To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:
Let \(f(x)=30{(2)}^x\). What is \(f(3)\)?
\[\begin{align*} f(x)&= 30{(2)}^x\\ f(3)&= 30{(2)}^3 \qquad \text{Substitute } x=3\\ &= 30(8) \qquad \text{Simplify the power first}\\ &= 240 \qquad \text{Multiply} \end{align*}\]
Remember to follow order of operations correctly. Exponent operations must be performed before multiplication.
\[f(3)=30{(2)}^3≠{60}^3=216,000 \nonumber\]
At the beginning of this section, we learned that the population of India was about \(1.25\) billion in the year 2013, with an annual growth rate of about \(1.2\%\). This situation is represented by the growth function \(P(t)=1.25{(1.012)}^t\), where \(t\) is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?
Solution
To estimate the population in 2031, we evaluate the models for \(t=18\), because 2031 is \(18\) years after 2013. Rounding to the nearest thousandth,
\[P(18)=1.25{(1.012)}^{18}≈1.549 \nonumber\]
There will be about \(1.549\) billion people in India in the year 2031.
Bismuth-210 is an isotope that radioactively decays by about 13% each day, meaning 13% of the remaining Bismuth-210 transforms into another atom (polonium-210 in this case) each day. If you begin with 100 mg of Bismuth-210, how much remains after one week?
Solution
With radioactive decay, instead of the quantity increasing at a percent rate, the quantity is decreasing at a percent rate. Our initial quantity is \(a\) = 100 mg, and our growth rate will be negative 13%, since we are decreasing: \(r\) = -0.13. This gives the equation:
\[Q(d)=100(1-0.13)^{d} =100(0.87)^{d}\nonumber\]
This can also be explained by recognizing that if 13% decays, then 87 % remains.
After one week, 7 days, the quantity remaining would be
\[Q(7)=100(0.87)^{7} =37.73\text{ mg of Bismuth-210 remains.}\nonumber \]
\(T(q)\) represents the total number of Android smart phone contracts, in thousands, held by a certain Verizon store region measured quarterly since January 1, 2016, interpret all the parts of the equation \(T(2)=86(1.64)^{2} =231.3056\).
Solution
Interpreting this from the basic exponential form, we know that 86 is our initial value. This means that on Jan. 1, 2016 this region had 86,000 Android smart phone contracts. Since \(b = 1 + r = 1.64\), we know that every quarter the number of smart phone contracts grows by 64%. \(T(2) = 231.3056\) means that in the \(2^{nd}\) quarter (or at the end of the second quarter) there were approximately 231,306 Android smart phone contracts.
Compound Interest
A certificate of deposit (CD) is a type of savings account offered by banks, typically offering a higher interest rate in return for a fixed length of time you will leave your money invested. Typically bank accounts and other savings instruments in which earnings are reinvested, such as mutual funds and retirement accounts, utilize compound interest. The term compounding comes from the behavior that interest is earned not on the original value, but on the accumulated value of the account.
The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing.
We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time \(t\), principal \(P\), annual percentage rate \(r\), and number of compounding periods in a year \(n\):
\[A(t)=P{\left (1+\dfrac{r}{n} \right )}^{nt} \nonumber\]
For example, observe Table \(\PageIndex{3}\), which shows the result of investing \($1,000\) at \(10\%\) APR for one year. Notice how the value of the account increases as the compounding frequency increases.
| Frequency | Value after \(1\) year |
|---|---|
| Annually | \($1100\) |
| Semiannually | \($1102.50\) |
| Quarterly | \($1103.81\) |
| Monthly | \($1104.71\) |
| Daily | \($1105.16\) |
Compound interest can be calculated using the formula
\[A(t)=P{\left (1+\dfrac{r}{n} \right )}^{nt} \nonumber\]
where
- \(A(t)\) is the account value,
- \(t\) is measured in years,
- \(P\) is the starting amount of the account, often called the principal, or more generally present value,
- \(r\) is the annual percentage rate (APR) expressed as a decimal, and
- \(n\) is the number of compounding periods in one year.
Note that sometime rates are given without stating that they are annual percentage rates. In most textbooks, including this one, it is implied that the rate is an annual percentage rate unless explicitly stated otherwise.
If you invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years?
Solution
Since we are starting with $3000, \(a = 3000\)
Our interest rate is 3%, so \(r = 0.03\)
Since we are compounding quarterly, we are compounding 4 times per year, so \(k = 4\)
We want to know the value of the account in 10 years, so we are looking for \(A(10)\), the value when \(t = 10\).
\[A(10)=3000\left(1+\dfrac{0.03}{4} \right)^{4(10)} =\$ 4045.05\nonumber\]
The account will be worth $4045.05 in 10 years.
(note that we round to the nearest cent when working with money)
A 529 plan is a college savings plan in which a relative can invest money to pay for a child’s later college tuition, and the account grows tax free. If Lily wants to set up a 529 account for her new granddaughter, wants the account to grow to $40,000 over 18 years, and she believes the account will earn 6% compounded semi-annually (twice a year), how much will Lily need to invest in the account now?
Solution
Since the account is earning 6%, \(r = 0.06\)
Since interest is compounded twice a year, \(k = 2\)
In this problem, we don’t know how much we are starting with, so we will be solving for a, the initial amount needed. We do know we want the end amount to be $40,000, so we will be looking for the value of a so that \(A(18) = 40,000\).
\[\begin{array}{l} {40,000=A(18)=a\left(1+\dfrac{0.06}{2} \right)^{2(18)} } \\ {40,000=a(2.8983)} \\ {a=\dfrac{40,000}{2.8983} \approx \$ 13,801} \end{array}\nonumber \]
Lily will need to invest $13,801 to have $40,000 in 18 years.
Euler's Number and The Natural Exponential Function
As we saw earlier, the amount we earn increases as we increase the compounding frequency. The table, though, 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 believe that although increasing the frequency of compounding will increase our result, there is an upper limit to this process.
To see this, let us examine the value of $1 invested at 100% interest for 1 year.
| Frequency | Value |
|---|---|
| Annual | $2 |
| Quarterly | $2.441406 |
| Monthly | $2.613035 |
| Daily | $2.714567 |
| Hourly | $2.718127 |
| Once per minute | $2.718279 |
| Once per second | $2.718282 |
These values do indeed appear to be approaching an upper limit. This value ends up being so important that it gets represented by its own letter, much like how \(\pi\) represents a number. This number is denoted by \(e\). The famous mathematician, Leonard Euler, was the first person to use this symbol to describe this number in 1731.
\(e\) is the letter used to represent the value that \(\left(1+\dfrac{1}{k} \right)^{k}\) approaches as \(k\) gets big.
\[e \approx 2.718282\nonumber\]
Because \(e\) is often used as the base of an exponential, most scientific and graphing calculators have a button that can calculate powers of \(e\), usually labeled \({e}^{x}\). Some computer software instead defines a function \(exp(x)\), where \(exp(x) = {e}^{x}\).
Because \(e\) arises when the time between compounds becomes very small, \(e\) allows us to define continuous growth and allows us to define a new toolkit function, \(f(x)=e^{x}\).
Continuous Growth can be calculated using the formula
\[f(x)=ae^{rx}\nonumber\]
where
\(a\) is the starting amount
\(r\) is the continuous growth rate
Euler's number is commonly used when describing quantities that change more or less continuously in nature, like chemical reactions, growth of large populations, and radioactive decay. The natural exponential function is an exponential function with a base \(e\).
The natural exponential function is the exponential function defined as \[f(x)=e^x \nonumber \]
Radon-222 decays at a continuous rate of 17.3% per day. How much will 100mg of Radon-222 decay to in 3 days?
Solution
Since we are given a continuous decay rate, we use the continuous growth formula. Since the substance is decaying, we know the growth rate will be negative: \(r\) = -0.173
\(f(3)=100e^{-0.173(3)} \approx 59.512\) mg of Radon-222 will remain.
Interpret the following: \(S(t)=20e^{0.12t}\) if \(S(t)\) represents the growth of a substance in grams, and time is measured in days.
- Answer
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An initial substance weighing 20g is growing at a continuous rate of 12% per day.
If $1000 is invested in an account earning 10% compounded continuously, find the value after 1 year.
Solution
Here, the continuous growth rate is 10%, so \(r\) = 0.10. We start with $1000, so \(a\) = 1000.
To find the value after 1 year,
\[f(1)=1000e^{0.10(1)} \approx \$ 1105.17\nonumber\]
Notice this is a $105.17 increase for the year. As a percent increase, this is \(\dfrac{105.17}{1000} =0.10517=10.517\%\) increase over the original $1000.