Unit 3, lesson 36

Exponential Functions and Growth/Decay

What an exponential function is, why its base must be positive and not 1, how equal steps in the input give equal ratios in the output, and how growth and decay differ in graph, range and asymptote.

  • Theory part 71
  • Practice part 72
  • Exponentials, e, and Logarithms

A culture of 100 bacteria that doubles every hour has 200 after one hour, 400 after two and 800 after three. Each hour adds a different amount, yet each hour multiplies the count by the same number. That steady multiplying is what makes a function exponential, and it is why such quantities look slow at first and then shoot up.

Key terms

Exponential function

Definition Let bb be a real number with b>0b > 0 and b≠1b \neq 1. The exponential function with base bb is

f(x)=bx,f(x) = b^x,

defined for every real number xx. For a rational exponent x=m/nx = m/n, with mm an integer and nn a positive integer,

bm/n=(bn)m,b^{m/n} = \bigl(\sqrt[n]{b}\bigr)^m,

and for an irrational xx the value bxb^x is the number that brb^r approaches as rational numbers rr approach xx. More generally, a function of the form f(x)=a⋅bxf(x) = a \cdot b^x with a constant a≠0a \neq 0 is also called exponential. Its initial value is f(0)=af(0) = a.

In plain words The variable sits in the exponent. That is the difference from a power function such as x2x^2, where the variable is the base and the exponent is fixed.

Example For f(x)=3⋅2xf(x) = 3 \cdot 2^x the initial value is 33 and each step of 11 in xx doubles the output.

xx−1-100112233
f(x)f(x)1.51.5336612122424

Inputs between the integers make sense too, for example f(1/2)=32≈4.243f(1/2) = 3\sqrt{2} \approx 4.243.

Positive base other than one

Definition The base bb of an exponential function must satisfy b>0b > 0 and b≠1b \neq 1. If b=1b = 1, then bx=1b^x = 1 for every xx, which is a constant function. If b=0b = 0, then bxb^x is 00 for every x>0x > 0 and undefined for every x≤0x \le 0. If b<0b < 0, then bxb^x is not a real number for many inputs, for instance whenever x=1/nx = 1/n with nn even, so it is not defined on all real numbers.

In plain words A base of 11 never changes anything, and a zero or negative base breaks down at fractional or negative inputs. A positive base other than 11 works for every real input.

Example With base −4-4, the input 1/21/2 would give

(−4)1/2=−4,(-4)^{1/2} = \sqrt{-4},

which is not a real number. With base 11, every input gives 1x=11^x = 1, so the graph is the flat line y=1y = 1 and nothing grows or shrinks.

Growth factor

Definition In f(x)=a⋅bxf(x) = a \cdot b^x the base bb is the growth factor per unit of input, because

f(x+1)=a⋅bx+1=b⋅f(x)for every real x.\begin{gathered} f(x + 1) = a \cdot b^{x + 1} = b \cdot f(x) \\[0.4em] \text{for every real } x. \end{gathered}

If b=1+rb = 1 + r, the number rr is the percent rate of change per unit of input, written as a decimal. When a>0a > 0, a positive rr means growth and −1<r<0-1 < r < 0 means decay. For 0<b<10 < b < 1 the base is often called the decay factor instead.

In plain words The growth factor is the number you multiply by each time the input goes up by 11. A 3 percent increase means multiplying by 1.031.03, and a 20 percent decrease means multiplying by 0.80.8.

Example A town of 80008000 people grows by 3 percent per year, so its growth factor is 1.031.03 and its population after tt years is

P(t)=8000⋅1.03t.P(t) = 8000 \cdot 1.03^t.

After one year P(1)=8240P(1) = 8240, and after ten years P(10)≈10,751P(10) \approx 10{,}751.

Equal input increments and equal output ratios

Definition For f(x)=a⋅bxf(x) = a \cdot b^x and any fixed step hh,

f(x+h)f(x)=a⋅bx+ha⋅bx=bhfor every real x.\begin{gathered} \frac{f(x + h)}{f(x)} = \frac{a \cdot b^{x + h}}{a \cdot b^x} = b^h \\[0.8em] \text{for every real } x. \end{gathered}

So equal input increments give equal output ratios. Conversely, suppose data points have equally spaced inputs x0,x0+h,x0+2h,…x_0, x_0 + h, x_0 + 2h, \ldots with h>0h > 0, and outputs y0,y1,y2,…y_0, y_1, y_2, \ldots with y0≠0y_0 \neq 0 and a constant ratio yk+1/yk=Ry_{k+1}/y_k = R, where R>0R > 0 and R≠1R \neq 1. Then all the points lie on the graph of the exponential function

f(x)=y0⋅R(x−x0)/h,f(x) = y_0 \cdot R^{(x - x_0)/h},

whose base is R1/hR^{1/h}. A linear function behaves differently, since equal input increments give it equal output differences.

In plain words To spot an exponential pattern in a table with evenly spaced inputs, divide each output by the one before. Constant ratios mean exponential, constant differences mean linear.

Example A bacteria count is measured every 3 hours.

Time tt in hours00336699
Count NN2002006006001800180054005400

The differences 400400, 12001200 and 36003600 are not equal, but every ratio is 33. So the count triples every 3 hours, and

N(t)=200⋅3t/3.N(t) = 200 \cdot 3^{t/3}.

The hourly growth factor is 31/3≈1.4423^{1/3} \approx 1.442, so after one hour there are about 288288 bacteria.

Exponential growth

Definition The function f(x)=a⋅bxf(x) = a \cdot b^x with a>0a > 0 and b>1b > 1 models exponential growth. It is increasing on the whole real line, f(x)→∞f(x) \to \infty as x→∞x \to \infty, and f(x)→0f(x) \to 0 as x→−∞x \to -\infty. Writing b=1+rb = 1 + r, the number r>0r > 0 is the growth rate per unit of input.

In plain words The quantity is multiplied by the same number bigger than 11 in each time step, so the amount added gets larger every step.

Example Suppose 10001000 dollars earns 5 percent interest, added once a year. The balance after tt full years is

A(t)=1000⋅1.05t dollars.A(t) = 1000 \cdot 1.05^t \text{ dollars.}
Years tt00101020203030
Balance A(t)A(t)1000.001000.001628.891628.892653.302653.304321.944321.94

Each 10 years multiplies the balance by 1.0510≈1.6291.05^{10} \approx 1.629. The first decade adds about 629629 dollars and the third adds almost 16691669.

Exponential decay

Definition The function f(x)=a⋅bxf(x) = a \cdot b^x with a>0a > 0 and 0<b<10 < b < 1 models exponential decay. It is decreasing on the whole real line, f(x)→0f(x) \to 0 as x→∞x \to \infty, and f(x)→∞f(x) \to \infty as x→−∞x \to -\infty. Writing b=1+rb = 1 + r as before, the rate rr satisfies −1<r<0-1 < r < 0, and the fraction lost per unit of input is −r=1−b-r = 1 - b. Since bx=(1/b)−xb^x = (1/b)^{-x}, the graph of a decay function is the mirror image in the yy-axis of a growth function with base 1/b>11/b > 1.

In plain words The quantity is multiplied by the same number between 00 and 11 in each time step. It keeps shrinking but never reaches zero.

Example Suppose the body removes 30 percent of a 200200 milligram dose of a medicine each hour. The amount left after tt hours is

M(t)=200⋅0.7t milligrams.M(t) = 200 \cdot 0.7^t \text{ milligrams.}

The values for t=0,1,2,3t = 0, 1, 2, 3 are 200200, 140140, 9898 and 68.668.6 milligrams. Each hour removes less than the hour before, because 30 percent of a smaller amount is a smaller amount.

Domain and range

Definition Let f(x)=a⋅bx+kf(x) = a \cdot b^x + k with a≠0a \neq 0, b>0b > 0 and b≠1b \neq 1. The domain of ff is the set of all real numbers. The function bxb^x takes every positive value exactly once and no other value, so the range of ff is

(k,∞) if a>0,(−∞,k) if a<0.\begin{gathered} (k, \infty) \text{ if } a > 0, \\[0.4em] (-\infty, k) \text{ if } a < 0. \end{gathered}

With k=0k = 0 this gives the range (0,∞)(0, \infty) for f(x)=a⋅bxf(x) = a \cdot b^x with a>0a > 0.

In plain words You can put in any real number. A positive base raised to any power is positive, so the outputs all lie on one side of the height kk and never reach it.

Example For g(x)=−2⋅3x+5g(x) = -2 \cdot 3^x + 5 the domain is all real numbers. Since 3x>03^x > 0, the term −2⋅3x-2 \cdot 3^x is always negative, so g(x)<5g(x) < 5. Every height below 55 is reached, since g(x)=yg(x) = y means 3x=(5−y)/23^x = (5 - y)/2, which has a solution whenever (5−y)/2>0(5 - y)/2 > 0. The range is (−∞,5)(-\infty, 5). For instance g(0)=3g(0) = 3 and g(2)=−13g(2) = -13.

Horizontal asymptote

Definition For f(x)=a⋅bx+kf(x) = a \cdot b^x + k with a≠0a \neq 0, b>0b > 0 and b≠1b \neq 1, the line y=ky = k is a horizontal asymptote. It is approached as x→−∞x \to -\infty when b>1b > 1 and as x→∞x \to \infty when 0<b<10 < b < 1. In the other direction ∣f(x)∣\lvert f(x) \rvert grows without bound, so there is no horizontal asymptote on that side. The graph never meets the line y=ky = k, because a⋅bx≠0a \cdot b^x \neq 0.

In plain words On one side the graph flattens out toward the height kk, getting closer and closer without touching it. On the other side it climbs or falls steeply.

Example Take f(x)=3⋅(1/2)x+1f(x) = 3 \cdot (1/2)^x + 1. Here 0<b<10 < b < 1, so the graph approaches y=1y = 1 as xx grows.

xx002244661010
f(x)f(x)441.751.751.18751.18751.0468751.0468751.0029…1.0029\ldots

The values stay above 11, because 3⋅(1/2)x>03 \cdot (1/2)^x > 0. To the left the graph rises quickly, for instance f(−3)=25f(-3) = 25.

Check yourself

  1. Is the data in this table exponential? If so, find a formula.
xx00112233
yy48483636272720.2520.25
Show the answer

The inputs are equally spaced and every ratio of consecutive outputs is 0.750.75, for example 36/48=0.7536/48 = 0.75. So the data is exponential with initial value 4848 and growth factor 0.750.75, that is y=48⋅0.75xy = 48 \cdot 0.75^x. This is decay by 25 percent per step.

  1. A culture of 500 bacteria doubles every 4 hours. Write a formula for the count after tt hours and find the count after 12 hours.
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Doubling every 4 hours gives N(t)=500⋅2t/4N(t) = 500 \cdot 2^{t/4}. After 12 hours the culture has doubled three times, and N(12)=500⋅23=4000N(12) = 500 \cdot 2^3 = 4000.

  1. Give the domain, the range and the horizontal asymptote of f(x)=6−4⋅(0.5)xf(x) = 6 - 4 \cdot (0.5)^x, and say from which side the graph approaches the asymptote.
Show the answer

The domain is all real numbers. Here a=−4<0a = -4 < 0 and k=6k = 6, so the range is (−∞,6)(-\infty, 6). The base 0.50.5 is between 00 and 11, so the horizontal asymptote y=6y = 6 is approached as x→∞x \to \infty. The term −4⋅(0.5)x-4 \cdot (0.5)^x is always negative, so the graph approaches y=6y = 6 from below.

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