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Week 13 · Chapter

Chapter 13 — Exponential Functions

College Algebra Free sample edition

The module's primary reading · ~40–50 minutes · Objective 8 (analyze exponential and logarithmic functions and their graphs, apply the properties of logarithms, and solve exponential and logarithmic equations and applied growth, decay, and compound-interest problems), Objective 3 (use function notation; determine the domain and range of a function; evaluate functions at numeric and algebraic inputs; and form sums, differences, products, quotients, and compositions of functions) · Read it first this week; the lecture, the tutorial, the quiz, and the Model/Build Studio all build on it.

In plain language

Some things change by adding the same amount each time, and some change by multiplying by the same number each time. This week is about the second kind. A function that multiplies is called an exponential function (a rule where the input sits up in the exponent, so every step multiplies the output by the same number). It describes savings that earn interest, a town that grows by a percent each year, a phone that loses value, and a sample that shrinks by half on a schedule. Once you can read the starting amount and the multiplier from the formula, you can predict any of them and check your answer on a graph.

Why you already care about this

Think about two things you may own right now: a phone and some savings. Say the phone cost 1,200 dollars and loses 25% of its value every year. After one year it is worth 900 dollars, and after two years it is worth 675 dollars. Now say 1,000 dollars sits in an account that pays 6% a year. After one year it holds 1,060.00 dollars, and after two years it holds 1,123.60 dollars.

One amount falls and the other rises, but both follow the same kind of rule. Each year you multiply by the same number: 0.75 for the phone and 1.06 for the savings. A city planner projecting a town's size uses that rule. So does a nurse tracking a medicine that leaves the body by half on a schedule, and so does a buyer comparing two savings offers.

In Week 12 you worked with roots and with exponents that are fractions. This week the variable itself moves up into the exponent, and that one change gives you a new family of functions.

By the end of this chapter you will be able to:

  • tell a table that adds the same amount from a table that multiplies by the same number;
  • read the initial value and the growth factor from a formula such as 3 · 2ˣ;
  • evaluate an exponential function at positive, zero, and negative inputs, doing the exponent first;
  • say whether an exponential function shows growth or decay, from its base alone;
  • read the y-intercept, the horizontal asymptote, and the range from the formula, and confirm them on a graph;
  • build a model from a percent rate, and compute compound-interest and continuously compounded balances.

Carry this sentence with you: lines add; exponentials multiply — and the base decides.

Where we are

  • Already covered: Week 1 set the exponent rules, including zero and negative exponents; Week 3 named functions, domain, and range; Week 4 read a line's constant rate of change; Week 10 met horizontal asymptotes; Week 12 used rational exponents.
  • This week: the exponential function f(x) = a · bˣ, its graph, growth and decay, percent-rate models, compound interest, and the natural base e.
  • How they connect: a line adds its slope at every step, and an exponential multiplies by its base, so Week 4's table and this week's table sit side by side.
  • Where we're going next: Week 14, Logarithmic Functions, runs this week's rule backwards, so you will later be able to ask which exponent gives a certain output.

Each earlier week hands you a tool. Week 1 gave you the powers of two, from 2⁰ = 1 up to 2¹⁰ = 1024. It also gave you two rules you need all week. Any nonzero base to the zero power is 1. A negative exponent means a reciprocal, so 2⁻² = 1 ⁄ 4, never a negative number. Week 3 taught you that f(x) names one output for one input, and that the range is the set of every output a function actually produces. Week 4 showed that a line's outputs change by the same amount at every step. Week 10 described a horizontal asymptote as a level line a graph gets closer and closer to at its far ends. This week every one of those returns.

1. Compare adding with multiplying

Picture two ways to fill a jar of coins. In the first, you drop in 3 coins every day. In the second, you double whatever is in the jar every day. On day one the two jars look alike, but they soon tell very different stories. One adds the same amount each time. The other multiplies by the same number each time.

You can see the difference in a table before you see any graph. Put the line from Week 4, g(x) = 3x + 4, beside a new function, f(x) = 3 · 2ˣ, at the same inputs.

  • A constant ratio is the pattern in a table where each output divided by the output before it gives the same number every time the input goes up by 1.

Worked example — adding versus multiplying. Compare the two functions at x = 0, 1, 2, 3.

  1. The line. g(x) = 3x + 4 at x = 0, 1, 2, 3 gives 4, 7, 10, 13.
  2. Subtract neighbors. 7 − 4 = 3, 10 − 7 = 3, 13 − 10 = 3. Every first difference is 3. You add 3 each time.
  3. The exponential. f(x) = 3 · 2ˣ at x = 0, 1, 2, 3 gives 3, 6, 12, 24.
  4. Divide neighbors. 6 ⁄ 3 = 2, 12 ⁄ 6 = 2, 24 ⁄ 12 = 2. Every ratio is 2. You double each time.
The line 3x + 4 and the exponential 3 · 2ˣ at the inputs 0 through 3
Input x Line 3x + 4 Exponential 3 · 2ˣ
0 4 3
1 7 6
2 10 12
3 13 24

In words: the line adds 3 at every step, so it has a constant difference. The exponential doubles at every step, so it has a constant ratio. At x = 0 the line is ahead, 4 to 3. By x = 3 the exponential is far ahead, 24 to 13.

To test a table, subtract neighbors and divide neighbors. If the differences match, you have a line. If the ratios match, you have an exponential.

Misconception check. Students often think any table that grows fast must be a line with a big slope. It is tempting because both columns go up. The cure: compute the differences. For 3, 6, 12, 24 they are 3, 6, 12, which change every time, so no single slope fits. The ratios are all 2, and a constant ratio is the sign of an exponential.

2. Read the initial value and the growth factor

Back to the jar that doubles. Two numbers tell its whole story: how many coins it starts with, and what you multiply by each day. An exponential function packs exactly those two numbers into its formula.

  • An exponential function is a function of the form f(x) = a · bˣ, with a ≠ 0, b > 0, and b ≠ 1, whose input sits in the exponent, so each one-unit step in the input multiplies the output by the same factor b.
  • The initial value is the number a, which is the output when the input is 0, because b⁰ = 1.
  • The growth factor is the base b, the number each output is multiplied by when the input goes up by 1.

Worked example — reading a and b. Name the initial value and the growth factor of f(x) = 3 · 2ˣ.

  1. The number in front. The initial value is a = 3. It is a coefficient that multiplies the power.
  2. The base with the exponent. The growth factor is b = 2. It is the base that carries the exponent x.
  3. Check against the table. The output at x = 0 is 3, and each output is 2 times the one before. Both match.

In words: 3 · 2ˣ starts at 3 and doubles with every step of 1 in x.

Why the rules on b? With b = 1, the outputs a · 1ˣ never change, so the function is flat, not exponential. The base must also be positive, so the rule works for every real input. A negative base would flip the sign of the output back and forth, and a base such as −2 cannot even be raised to the power 1 ⁄ 2 as a real number.

Misconception check. Students often swap the two numbers and say 3 · 2ˣ "starts at 2 and triples." It is tempting because both numbers sit side by side. The cure: the base is the number wearing the exponent. Here the exponent x sits on the 2, so 2 is the growth factor, and the 3 out front is the initial value. Check it at x = 0: 3 · 2⁰ = 3 · 1 = 3, so the start is 3.

3. Work the exponent first: positive, zero, and negative inputs

Here is where the Week 1 order of operations pays off. In 3 · 2ˣ, the exponent belongs to the 2 alone. So you raise 2 to the power first, and only then multiply by 3.

Worked example — a positive input. Evaluate f(2) for f(x) = 3 · 2ˣ.

  1. Substitute. f(2) = 3 · 2².
  2. Exponent first. 2² = 4.
  3. Then multiply. 3 · 4 = 12.

In words: at x = 2 the function equals 12.

Worked example — zero. Evaluate f(0).

  1. Substitute. f(0) = 3 · 2⁰.
  2. Zero exponent. Any nonzero base to the zero power is 1, so 2⁰ = 1.
  3. Multiply. 3 · 1 = 3.

In words: the output at x = 0 is the initial value, 3.

Worked example — negative inputs. Evaluate f(−2) and f(−3).

  1. Substitute. f(−2) = 3 · 2⁻².
  2. Negative exponent means reciprocal. 2⁻² = 1 ⁄ 2² = 1 ⁄ 4.
  3. Multiply. 3 · (1 ⁄ 4) = 3 ⁄ 4 = 0.75.
  4. One more. f(−3) = 3 · 2⁻³ = 3 · (1 ⁄ 8) = 3 ⁄ 8 = 0.375. The 8 is the 2³ = 8 from the powers of two.

In words: at x = −2 the output is 0.75, and at x = −3 it is 0.375. Both are small, and both are positive.

Here is the full table for the week's function.

Outputs of f(x) = 3 · 2ˣ for the inputs −3 through 3
Input x Output f(x)
−3 0.375
−2 0.75
−1 1.5
0 3
1 6
2 12
3 24

Each output is double the one before. The seven outputs sum to 47.625.

Misconception check — the 6ˣ trap. Students often multiply 3 · 2 first and write 3 · 2ˣ as 6ˣ. It is tempting because 3 and 2 sit right next to each other. Test it at x = 2: 6² = 36, versus 3 · 2² = 12, a ratio of 3 to 1, so the wrong reading is three times too big. Since 36 ≠ 12, 3 · 2ˣ is not 6ˣ. The cure is the order of operations: exponent first, then multiply.

Misconception check — the negative output trap. Students often think a negative input gives a negative output. It is tempting because the minus sign is right there. But a negative exponent means a reciprocal, never a negative number. f(−2) = 0.75 is small but positive.

In practice (your tools). In Desmos, type f(x) = 3 * 2^x on one line and f(-2) on the next. Desmos shows 0.75. Then type 6^2 and 3 * 2^2 on two more lines to see 36 and 12 side by side. If you ever get a negative output from 3 · 2ˣ, check the parentheses you typed.

4. Use the base to tell growth from decay

Now flip the story. Picture a 100-gram sample of a substance that loses half of its mass every period. After one period, 50 grams remain. After two, 25 grams remain. Each step multiplies by the same number, 1 ⁄ 2, so this is also an exponential function. It shrinks instead of growing.

  • Exponential growth is the behavior of f(x) = a · bˣ with a > 0 and a base b greater than 1, where the outputs get larger by the same factor at every step.
  • Exponential decay is the behavior of f(x) = a · bˣ with a > 0 and a base b between 0 and 1, where the outputs shrink by the same factor at every step.
  • A decay factor is a base b between 0 and 1, the fraction of the amount that remains after each one-unit step.
  • A half-life is the length of time it takes a decaying amount to fall to half of its size.

Worked example — the halving sample. The sample follows A(t) = 100 · (1 ⁄ 2)ᵗ, where t counts half-lives. Find A(3) and classify the function.

  1. Read the parts. The initial value is 100 grams. The decay factor is b = 1 ⁄ 2, which is between 0 and 1, so this is exponential decay.
  2. Substitute. A(3) = 100 · (1 ⁄ 2)³.
  3. Exponent first. (1 ⁄ 2)³ = 1 ⁄ 8.
  4. Multiply. 100 · (1 ⁄ 8) = 100 ⁄ 8 = 12.5.

In words: after three half-lives, 12.5 grams remain.

Mass of the 100-gram sample, in grams, after 0 through 4 half-lives
Half-lives t Mass A(t) in grams
0 100
1 50
2 25
3 12.5
4 6.25

Check the table: each mass is half the one before, since 50 ⁄ 100 = 1 ⁄ 2 and 6.25 ⁄ 12.5 = 1 ⁄ 2. The five masses sum to 193.75 grams.

Now look back at 3 · 2ˣ. Its base, 2, is greater than 1, so it shows exponential growth everywhere, even on the left side of the table where the outputs are small.

Misconception check. Students often see f(−2) = 0.75 and call 3 · 2ˣ a decay function, because 0.75 is small. It is tempting because small numbers feel like shrinking. The cure: growth or decay is decided by the base, not by the size of one output. Read the table left to right: 0.375, 0.75, 1.5, 3, and onward. The outputs grow at every step because b = 2 is greater than 1. The base decides.

5. Read the graph from the formula

A graph is a picture of the whole table at once. For an exponential, three features come straight from the formula, and you can confirm each one on a screen.

The y-intercept. Set x = 0. You get f(0) = 3, so the graph of 3 · 2ˣ crosses the y-axis at (0, 3). The sample's graph crosses at (0, 100). In each case the point is (0, a), the initial value.

The horizontal asymptote. Walk left through the table: 3, 1.5, 0.75, 0.375. The outputs keep halving. They get closer to 0 but never reach it, because half of a positive number is still positive. So the x-axis, y = 0, is a horizontal asymptote, a horizontal line y = c that a graph approaches at its far left or far right.

Always rising or always falling. The graph of 3 · 2ˣ rises from left to right everywhere because b = 2 > 1. The graph of 100 · (1 ⁄ 2)ᵗ falls everywhere because the base 1 ⁄ 2 is between 0 and 1. Neither graph turns around, so you never see a peak or a valley.

Worked example — domain and range. State the domain and range of f(x) = 3 · 2ˣ.

  1. Domain. You can raise 2 to any real power, so every input works. The domain is all real numbers, (−∞, ∞).
  2. Sign of the outputs. 2ˣ is positive for every input, and a = 3 is positive, so every output 3 · 2ˣ is positive.
  3. Range. The outputs cover every positive number but never 0 or a negative. The range is (0, ∞).

In words: the graph lives entirely above the x-axis. It never touches it and never crosses it.

In practice (your tools). In Desmos or GeoGebra, type y = 3 * 2^x and y = 3x + 4 on one screen. Click the curve where it meets the y-axis to see the point (0, 3). Zoom out to the left and watch the curve flatten toward the x-axis without reaching it. The two graphs also cross: the exponential starts below the line at x = 0, where 3 < 4, and passes it before x = 3, where 24 > 13.

Misconception check. Students often sketch an exponential that dips through the x-axis on the left. It is tempting because many graphs in this course, such as lines and parabolas, do cross it. The cure: compute an output far to the left. f(−3) = 0.375, and f(−10) would be 3 ⁄ 1024, still positive. With a > 0, every output is positive, so the curve hugs the x-axis without ever touching it.

One honesty note: this picture assumes a > 0. If a were negative, every output would be negative, and the graph would be flipped below the x-axis. The functions you meet this week all start with a positive amount.

6. Turn a percent rate into a factor

Real growth is usually stated as a percent. A town grows 3% a year. A phone loses 25% of its value each year. To use the exponential formula, you turn the percent into a factor.

  • A percent rate r is written as a decimal, so 3% is r = 0.03 and 25% is r = 0.25.
  • Growth at rate r multiplies by b = 1 + r, because you keep the whole amount and add the percent.
  • Decay at rate r multiplies by b = 1 − r, because you keep the whole amount and lose the percent.

Worked example — a growing town. A town of 800 people grows 3% per year. Write the model and estimate the population after 10 years.

  1. Rate to factor. r = 0.03, so b = 1 + 0.03 = 1.03.
  2. The model. P(t) = 800 · (1.03)ᵗ, with t in years.
  3. Ten years. P(10) = 800 · (1.03)¹⁰.
  4. Use the pinned value. With the given pinned constant (1.03)¹⁰ ≈ 1.3439, you get 800 · (1.03)¹⁰ ≈ 1,075 people.

In words: after ten years of 3% growth the town has about 1,075 people. A planner would use that number to decide whether the town needs another school bus route.

Worked example — a phone that loses value. A 1,200-dollar phone loses 25% of its value each year. Write the model and find its value after one and two years.

  1. Rate to factor. r = 0.25, so b = 1 − 0.25 = 0.75.
  2. The model. V(t) = 1200 · (0.75)ᵗ, with t in years.
  3. One year. V(1) = 1200 · 0.75 = 900, so it is worth 900 dollars after one year.
  4. Two years. V(2) = 900 · 0.75 = 675, so it is worth 675 dollars after two years.

In words: the phone keeps 75% of its value each year. A seller would use this to decide when trading it in still makes sense.

In practice (your tools). In Desmos, type y = 800 * 1.03^x and click the curve at x = 10; you see a value a little above 1,075. Then type y = 1200 * 0.75^x and check the points (1, 900) and (2, 675).

Misconception check. Students often write 3% growth as b = 0.03 or as b = 3. It is tempting because 3 and 0.03 are the numbers in the problem. The cure: b = 0.03 would shrink the town to 3% of its size each year, and b = 3 would triple it. The factor is 1 + r = 1.03. You add the rate to 1.

One caveat: no real town grows at exactly 3% forever. The model is a tool for a short horizon, and a planner updates it as new counts come in.

7. Compare compound and simple interest

Money in a savings account is the classic exponential. The bank adds interest, and then the next round of interest is paid on the bigger balance.

  • The principal P is the amount of money first deposited or borrowed, before any interest is added.
  • Simple interest is interest paid only on the original principal, the same dollar amount every year.
  • Compound interest is interest that is added to the balance at the end of each period, so later interest is earned on earlier interest too.
  • A compounding period is the length of time between the moments when interest is added; n counts how many periods fit in one year.

The compound-interest formula is A = P(1 + r ⁄ n)^(nt). Here A is the balance in dollars and r is the annual rate as a decimal. The letter n is the number of compounding periods per year, and t is the time in years.

Worked example — annual compounding. 1,000 dollars earns 6% compounded once a year. Find the balance after one and two years.

  1. The parts. P = 1000, r = 0.06, n = 1, so A = 1000 · (1.06)ᵗ. The growth factor is 1.06. You multiply by 1.06 each year.
  2. One year. A(1) = 1000 · 1.06 = 1,060.00 dollars.
  3. Two years. A(2) = 1060 · 1.06 = 1,123.60 dollars.
  4. Check with the formula. That matches 1000 · (1.06)² = 1,123.60 dollars.

In words: after two years the account holds 1,123.60 dollars.

Balance of 1,000 dollars at 6% compounded annually, in dollars, at the end of years 0 through 5
Year t Balance in dollars
0 1,000.00
1 1,060.00
2 1,123.60
3 1,191.02
4 1,262.48
5 1,338.23

Each balance is rounded to the cent, and the six balances sum to 6,975.33 dollars.

Worked example — compound versus simple. How much more does compounding earn in two years?

  1. Simple interest. Simple interest pays 6% of the original 1,000 dollars each year, which is 60 dollars. After two years, 1000 · (1 + 0.06 · 2) = 1000 · 1.12 = 1,120.00 dollars.
  2. Subtract. 1,123.60 dollars − 1,120.00 dollars = 3.60 dollars.
  3. Explain it. Compounding earned 3.60 dollars more because the first year's 60 dollars of interest earned 6% in year two, and 60 · 0.06 = 3.60.

In words: compounding pays interest on interest. Over two years that is 3.60 dollars; over longer stretches the gap grows much wider, which is why a saver compares how often a bank compounds.

Worked example — twice a year. 2,000 dollars earns 4% compounded twice a year (n = 2) for t = 1 year.

  1. Rate per period. Each period pays r ⁄ n = 0.04 ⁄ 2 = 0.02.
  2. Number of periods. There are nt = 2 · 1 = 2 periods, so A = 2000 · (1.02)².
  3. First half-year. 2000 · 1.02 = 2,040.00 dollars.
  4. Second half-year. 2040 · 1.02 = 2,080.80 dollars.
  5. Compare. Compounded once a year it would be 2000 · 1.04 = 2,080.00 dollars, so twice a year earns 80 cents more.

In words: more compounding periods mean slightly more money, because interest starts earning interest sooner.

In practice (your tools). On a calculator, type 1000*(1+0.06/1)^(1*2) and you see 1123.6. For the twice-a-year account, type 2000*(1+0.04/2)^(2*1) and you see 2080.8. Put the whole exponent in parentheses, or the calculator multiplies at the wrong time.

Misconception check. Students often compute compound interest the simple way, adding 60 dollars a year, and get 1,120.00 dollars instead of 1,123.60 dollars. It is tempting because adding the same amount each year feels natural. The cure: compound interest is a multiplying story, not an adding story. Each year you multiply the current balance by 1.06, so year two's interest is 6% of 1,060 dollars, not 6% of 1,000 dollars.

8. Find the natural base e and compound continuously

What happens if a bank compounds more and more often: every month, every day, every second? The balance grows a little each time, but it does not run off to infinity. It levels off. The number it levels off at is one of the most important constants in mathematics.

Worked example — toward e. Compute (1 + 1 ⁄ n)ⁿ for larger and larger n.

  1. n = 1. (1 + 1 ⁄ 1)¹ = 2¹ = 2.
  2. n = 2. (1 + 1 ⁄ 2)² = 1.5² = 2.25.
  3. Bigger n. Rounded to two decimals, n = 12 gives 2.61, n = 365 gives 2.71, and n = 1,000,000 gives 2.72.
Values of (1 + 1 ⁄ n)ⁿ for growing n, rounded to two decimals after n = 2
n (1 + 1 ⁄ n)ⁿ
1 2
2 2.25
12 2.61
365 2.71
1,000,000 2.72

In words: the values climb, but they level off. The number they level off at is the constant e ≈ 2.71828.

  • The natural base e is the irrational constant e ≈ 2.71828 that (1 + 1 ⁄ n)ⁿ levels off toward as n grows.
  • Continuous compounding is compound interest added at every instant, so a balance follows A = Pe^(rt).

Like π, e is irrational, so its decimal never ends or repeats. You use the e key on your calculator. You never type 2.7 by hand.

Worked example — continuous compounding. 1,500 dollars earns 5% compounded continuously for 4 years.

  1. The exponent. rt = 0.05 · 4 = 0.2, so A = 1500 · e^0.2.
  2. The pinned value. You use the given pinned constant e^0.2 ≈ 1.2214.
  3. Multiply. Then A ≈ 1500 · 1.2214 ≈ 1,832.10 dollars.

In words: after four years of continuous compounding the account holds about 1,832.10 dollars.

Growth and decay with e. Any continuous model A = A₀ · e^(kt) starts at A₀ and grows when k > 0 and decays when k < 0. In A = Pe^(rt), A₀ = P and k = r = 0.05, so it grows. In A = A₀ · e^(−0.15t), k = −0.15, so it decays: the minus sign sits inside k itself. You only need to read the sign of k this week.

In practice (your tools). In Desmos, type 1500*e^(0.05*4) and you see about 1832.10. On a handheld calculator, look for the e^x key, usually above the ln key, and enter 0.2 as the exponent.

Misconception check. Students often treat e as a variable to solve for, or round it to 2.7 or 3 in the middle of a problem. It is tempting because e is a letter. The cure: e is a fixed number, like π, and it never changes. Use the calculator's e key and round only at the end, to the cent for money.

A reflection to choose from. Pick one option, (a) or (b), and write two or three sentences.

  • (a) Think of something in your own life that grows or shrinks by a percent, such as a price, a following, or a balance. Describe its starting amount and its factor, and say whether it shows growth or decay.
  • (b) Pick the trap from this chapter you are most likely to fall into: reading 3 · 2ˣ as 6ˣ, calling a small output decay, or adding interest instead of multiplying. Describe a short habit that would catch it before it costs you points.

Key terms

  • Constant ratio — The pattern in a table where each output divided by the output before it gives the same number every time the input goes up by 1, as in 3, 6, 12, 24, where every ratio is 2; it is the signature of an exponential function. (first used: 1. Compare adding with multiplying)
  • Exponential function — A function of the form f(x) = a · bˣ, with a ≠ 0, b > 0, and b ≠ 1, whose input sits in the exponent, so each one-unit step in the input multiplies the output by the same factor b. (first used: 2. Read the initial value and the growth factor)
  • Initial value — The number a in f(x) = a · bˣ, which is the output when the input is 0, because b⁰ = 1; on the graph it is the height where the curve crosses the y-axis. (first used: 2. Read the initial value and the growth factor)
  • Growth factor — The base b in f(x) = a · bˣ, the number each output is multiplied by when the input goes up by 1; for growth at a percent rate r it equals 1 + r. (first used: 2. Read the initial value and the growth factor)
  • Exponential growth — The behavior of f(x) = a · bˣ with a > 0 and a base b greater than 1, where the outputs get larger by the same factor at every step and the graph rises from left to right everywhere. (first used: 4. Use the base to tell growth from decay)
  • Exponential decay — The behavior of f(x) = a · bˣ with a > 0 and a base b between 0 and 1, where the outputs shrink by the same factor at every step and the graph falls from left to right toward the x-axis without reaching it. (first used: 4. Use the base to tell growth from decay)
  • Decay factor — A base b between 0 and 1 in f(x) = a · bˣ, the fraction of the amount that remains after each one-unit step; for decay at a percent rate r it equals 1 − r. (first used: 4. Use the base to tell growth from decay)
  • Half-life — The length of time it takes a decaying amount to fall to half of its size, so a sample measured in half-lives follows A(t) = A₀ · (1 ⁄ 2)ᵗ. (first used: 4. Use the base to tell growth from decay)
  • Principal — The amount of money first deposited or borrowed, written P in an interest formula, before any interest is added. (first used: 7. Compare compound and simple interest)
  • Simple interest — Interest paid only on the original principal, the same dollar amount every year, so a balance after t years is P(1 + rt). (first used: 7. Compare compound and simple interest)
  • Compound interest — Interest that is added to the balance at the end of each period so that later interest is earned on earlier interest too; the balance after t years is A = P(1 + r ⁄ n)^(nt). (first used: 7. Compare compound and simple interest)
  • Compounding period — The length of time between the moments when interest is added to a balance; n counts how many compounding periods fit in one year, so n = 1 is annual and n = 2 is twice a year. (first used: 7. Compare compound and simple interest)
  • Natural base e — The irrational constant e ≈ 2.71828 that (1 + 1 ⁄ n)ⁿ levels off toward as n grows; it is the base used for continuous growth and decay. (first used: 8. Find the natural base e and compound continuously)
  • Continuous compounding — Compound interest added at every instant instead of at separate periods, so a balance follows A = Pe^(rt), with r the annual rate as a decimal and t the time in years. (first used: 8. Find the natural base e and compound continuously)

Summary

  • A line adds the same amount each step; an exponential multiplies by the same factor.
  • In a · bˣ, a is the starting output and b is the multiplier.
  • Do the exponent first: 3 · 2ˣ is never 6ˣ.
  • A negative input gives a small positive output, never a negative one.
  • A base above 1 means growth; a base between 0 and 1 means decay.
  • The graph crosses the y-axis at (0, a) and hugs the x-axis without touching it.
  • Growth at rate r uses 1 + r; decay at rate r uses 1 − r.
  • Compound interest multiplies each period; simple interest adds the same amount.
  • e is a fixed irrational number, about 2.71828, used for continuous compounding.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

  1. For h(x) = 2 · 5ˣ, find h(2) and h(−1). Then explain why h(2) is not 100.
  2. Is k(x) = 64 · (3 ⁄ 4)ˣ growth or decay? Give its y-intercept, find k(2), and name its horizontal asymptote.
  3. A coffee shop's loyalty app has 2,000 members, and membership grows 10% per month. Write a model and find the number of members after 2 months.
  4. A deposit of 4,000 dollars earns 10% compounded twice a year. Find the balance after one year, and compare it with interest compounded once a year.

Answers (worked reasoning — read after attempting)

  1. Exponent first. h(2) = 2 · 5² = 2 · 25 = 50. For the negative input, 5⁻¹ = 1 ⁄ 5, so h(−1) = 2 · (1 ⁄ 5) = 2 ⁄ 5 = 0.4, small but positive. The value 100 comes from multiplying 2 · 5 = 10 first and then squaring, 10² = 100; that reads 2 · 5ˣ as 10ˣ, which breaks the order of operations.
  2. The base 3 ⁄ 4 is between 0 and 1, so this is decay. The y-intercept is (0, 64), the initial value. k(2) = 64 · (3 ⁄ 4)² = 64 · (9 ⁄ 16) = 576 ⁄ 16 = 36. The horizontal asymptote is the x-axis, y = 0: the outputs shrink toward 0 but stay positive.
  3. A 10% growth rate gives r = 0.10 and b = 1 + 0.10 = 1.10, so M(t) = 2000 · (1.10)ᵗ with t in months. After one month, 2000 · 1.10 = 2,200 members. After two months, 2200 · 1.10 = 2,420 members. Writing b = 0.10 would wrongly shrink the app to a tenth of its size each month.
  4. Each half-year pays r ⁄ n = 0.10 ⁄ 2 = 0.05, and there are nt = 2 · 1 = 2 periods. After the first half-year, 4000 · 1.05 = 4,200 dollars. After the second, 4200 · 1.05 = 4,410 dollars. Compounded once a year, the balance would be 4000 · 1.10 = 4,400 dollars, so compounding twice a year earns 10 dollars more, because the first half-year's 200 dollars of interest earns 5% in the second half.

Next up in this module: the Readings & Resources page, the lecture tutorial, the practice set, the quiz, and the discussion — and next week the logarithm gives you the rule that runs an exponential backwards.

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