The number e as the limit of one plus one over n raised to the n, simplifying and evaluating natural base expressions, graphing y equals a times e to the rx, translating and modelling with natural base functions, and continuously compounded interest.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 7 — Exponential and Logarithmic Functions
Use Functions Involving e
Objectives
Five outcomes. The first explains where a strange-looking constant comes from.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 492-497 — the lesson these objectives are drawn from
Warm-up
Lesson 7.1 left a question hanging: compounding more often gives more, but how much more?
Discussion prompt
Compute one plus one over n, raised to the n, for n equal to 10, 100 and 1000. What is happening?
Hint: The base shrinks toward 1 while the exponent grows.
Answer:
\[ n=10: \; 2.59374 \qquad n=100: \; 2.70481 \qquad n=1000: \; 2.71692 \]
The two effects nearly cancel, and the values settle rather than running away. Their limit is an irrational number called e, about 2.718281828, and it turns out to be the natural base for every continuous growth or decay process.
Concept
The number e is defined as the limit of one plus one over n, raised to the n, as n grows without bound. It is irrational, like pi, and it is the base that describes growth happening continuously rather than in discrete steps.
natural base e — The irrational number approximately equal to 2.718281828, defined as the limit of one plus one over n raised to the n as n approaches infinity. It is named after Leonhard Euler.
\[ \left(1+\tfrac{1}{n}\right)^n \to e \approx 2.718281828 \]
Nothing about the algebra of e is new. It is a number greater than 1, so every rule and graph shape from Lessons 5.1, 7.1 and 7.2 applies to it unchanged.
Figure (svg): A table showing the quantity one plus one over n, raised to the n, converging to the number e
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 492-492
Section
Section 1
Concept
As n increases, one plus one over n gets closer to 1 while the exponent n grows without bound. The two effects balance, and the values converge to about 2.71828 rather than growing without limit or collapsing to 1.
\[ \left(1+\tfrac{1}{n}\right)^n, \quad n = 10, 100, 1000, \dots \]
Euler discovered the constant, and it joins pi and the imaginary unit among the numbers important enough to be given their own symbol.
Figure (svg): A table showing the quantity one plus one over n, raised to the n, converging to the number e
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 492-492 — The Natural Base e
Picture it
The table from the textbook, with the last three highlighted.
Figure (svg): A table showing the quantity one plus one over n, raised to the n, converging to the number e
From n equal to ten thousand onward the first five digits stop changing. The number is irrational, so the digits never repeat, but they do settle one at a time.
Worked example
The two competing effects, examined.
\[ \text{Explain why } \left(1+\tfrac{1}{n}\right)^n \text{ neither grows without bound nor collapses to } 1. \]
Look at the base alone
Why: One plus one over n approaches 1 as n grows, since one over n shrinks to zero.
\[ \text{base to } 1 \]
Look at the exponent alone
Why: The exponent n grows without bound.
Note the competition
Why: A base fixed above 1 raised to a growing power would explode; a base of exactly 1 raised to any power stays 1.
Read the table
Why: The values rise but by less and less, settling near 2.71828.
Figure (svg): The solution to Worked example watch the limit form shown as a ladder of expressions, one row per algebraic move
\[ \left(1+\tfrac{1}{n}\right)^n \to e \]
Verify: test the two extremes for contrast
Why: With a fixed base of 1.1, raising to the power n gives 2.6, 13,781 and then astronomically more — no limit at all. With a base of exactly 1 the value is always 1. The limit form sits precisely between these, which is why it produces a finite number bigger than 1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 492-492
Sorting
e is about 2.718.
Sort into buckets
Sort each number.
Knowing that e sits between 2.7 and 2.72 lets you estimate any expression involving it before reaching for a calculator.
Worked example
Where e sits among the numbers with symbols.
\[ \text{Compare } e \text{ with } \pi \text{ and } i \text{ as special numbers.} \]
Recall pi
Why: It is irrational, about 3.14159, and arises from the ratio of a circle's circumference to its diameter.
Recall i
Why: It was invented in Lesson 4.6 so that every quadratic would have solutions.
Place e
Why: It is irrational, about 2.71828, and arises as the limit of a growth process.
Note what they share
Why: Each is given a symbol because writing its decimal expansion is impossible and its value comes up constantly.
Figure (svg): The solution to Worked example e beside the other named constants shown as a ladder of expressions, one row per algebraic move
\[ e \approx 2.718281828 \]
Verify: check that e is genuinely between 2 and 3
Why: Every value in the table lies between 2.5 and 2.72, so the limit is comfortably between 2 and 3. That means e to the x behaves like a growth function between 2 to the x and 3 to the x — closer to the first — which is a useful way to sanity-check any calculation involving it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 492-492
Trap
\[ e^2 \cdot e^5 \]
Leave e as an unknown and collect nothing
Why: The letter is taken to stand for something not yet known.
\[ \text{cannot be simplified} \quad \text{(wrong)} \]
The two exponents can be added exactly as for base 2 or base 10, because e is a specific number.
\[ e^2 \cdot e^5 = e^7 \approx 1096.6 \]
Treat e as a number greater than 1
Why: It is a constant, so every exponent property applies to it.
\[ e \approx 2.718 \text{, a fixed number} \]
A letter used as a constant behaves like a number, not like a variable. The same is true of pi, which nobody hesitates to combine.
Fill the middle
The table.
Fill in the blanks
n = 12.71828000___000: \; \left(1+\tfrac______\right)^n \approx ___
Why: At a million the value agrees with e to five decimal places. The convergence is slow — each extra digit of accuracy needs roughly ten times more n — which is why the limit is worth naming rather than recomputing.
Prediction
Commit before reasoning.
Predict first
One point one raised to the n grows without bound. Why does one plus one over n, raised to the n, not?
Correct: Because the base shrinks toward 1 as the exponent grows.
\[ 1.1^{100} \approx 13{,}781 \quad \text{but} \quad (1.01)^{100} \approx 2.705 \]
Why: In 1.1 to the n the base is fixed, so every extra factor multiplies by the same 1.1 and the product grows without limit. Here the base is 1 plus one over n, which gets closer to 1 as n grows, so each new factor contributes less. The two effects balance exactly, and the balance point is e.
Matching
Each named number came from somewhere.
Match the pairs
Why: Three of the four are irrational real numbers and one is imaginary. Each was named because it turns up constantly and cannot be written exactly as a fraction or a decimal, which is exactly why symbols for them exist.
Section
Section 2
Concept
Every exponent property from Lesson 5.1 applies to expressions containing e: powers of e multiply by adding exponents, divide by subtracting, and a power of a power multiplies them. A calculator's e key evaluates any of them.
\[ e^2\cdot e^5 = e^7, \quad \frac{12e^4}{3e^3} = 4e, \quad (5e^{-3x})^2 = 25e^{-6x} \]
A simplified answer carries positive exponents, so 25 e to the negative 6x is written as 25 over e to the 6x — the same convention as everywhere since Lesson 5.1.
Figure (svg): Three natural base expressions simplified by the exponent properties
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 492-492 — Simplify natural base expressions
Picture it
Example 1, all three parts, each naming its property.
Figure (svg): Three natural base expressions simplified by the exponent properties
Nothing in any of the three is specific to e. Replacing e with 2 or with 10 would produce identical working.
Worked example
Example 1, all three parts.
\[ \text{Simplify } e^2e^5, \; \tfrac{12e^4}{3e^3}, \; (5e^{-3x})^2. \]
First: add the exponents
Why: Two plus 5 is 7.
\[ e ^{7} \]
Second: divide coefficients and subtract exponents
Why: Twelve over 3 is 4, and 4 minus 3 is 1.
\[ 4 e \]
Third: distribute the outer power
Why: Five squared is 25, and negative 3x times 2 is negative 6x.
\[ 25 e ^{-6 x} \]
Third: clear the negative exponent
Why: A negative exponent becomes a reciprocal.
\[ 25 / e ^{6 x} \]
Figure (svg): The solution to Worked example simplify three expressions shown as a ladder of expressions, one row per algebraic move
\[ e^7, \quad 4e, \quad \tfrac{25}{e^{6x}} \]
Verify: check one numerically
Why: The second: 12 times e to the fourth is about 655.2, and 3 times e cubed is about 60.3; their quotient is about 10.87, which is 4 times e, or 4 times 2.718. The algebra and the arithmetic agree, as they must when e is treated as a number.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 492-492
Matching
Apply the matching property.
Match the pairs
Why: In the last two the coefficient is raised to the outer power as well as the base, which is the step most often half-done: 5 squared is 25 and 10 cubed is 1000, not 5 and 10.
Worked example
Guided Practice 1 to 5.
\[ \text{Simplify } e^7e^4, \; 2e^{-3}\cdot 6e^5, \; \tfrac{24e^8}{4e^5}, \; (10e^{-4x})^3, \text{ and evaluate } e^{3/4}. \]
First: add
Why: Seven plus 4 is 11.
\[ e ^{11} \]
Second: multiply coefficients, add exponents
Why: Two times 6 is 12, and negative 3 plus 5 is 2.
\[ 12 e ^{2} \]
Third: divide and subtract
Why: Twenty-four over 4 is 6, and 8 minus 5 is 3.
\[ 6 e ^{3} \]
Fourth: distribute the cube
Why: Ten cubed is 1000, and negative 4x times 3 is negative 12x.
\[ 1000 / e ^{12 x} \]
Fifth: use the calculator's e key
Why: E to the three quarters is about 2.117.
\[ \text{about } 2.12 \]
Figure (svg): The solution to Worked example four more, plus an evaluation shown as a ladder of expressions, one row per algebraic move
\[ e^{11}, \; 12e^2, \; 6e^3, \; \tfrac{1000}{e^{12x}}, \; 2.117 \]
Verify: sanity-check the evaluation
Why: Three quarters is between 0 and 1, so e to the three quarters lies between e to the zero, which is 1, and e itself, which is 2.718. The value 2.117 is in that range and closer to the top, which fits an exponent of 0.75. Estimating the range before reading a display catches a mistyped exponent.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 493-493
Error analysis
A student simplifies a product of powers of e.
Annotate
On: \( e^2 \cdot e^5 = e^{10} \)
Exactly the error that Lesson 5.1 warned about, with e in place of a numeral. Writing out a small case — e times e, times e times e times e — settles it.
Fill the middle
Guided Practice 2.
Fill in the blanks
2e^2\cdot 6e^___ = 12e^___ = 12e^___}
Why: Negative 3 plus 5 is 2, and the coefficients 2 and 6 multiply to 12. The coefficient and the exponent are handled separately, which is the pattern for every product of this kind.
Sorting
Read the structure.
Sort into buckets
Sort each expression by the first property to apply.
Five expressions, five different first moves — and every one of the five rules was learned in Lesson 5.1 with numerals rather than e.
Prediction
Commit before reasoning.
Predict first
Can e to the x be zero or negative for some real x?
Correct: No — e is positive, so every power of it is positive.
\[ e^{-10} = \tfrac{1}{e^{10}} \approx 0.0000454 > 0 \]
Why: E is about 2.718, a positive number, and a positive base raised to any real power stays positive. At x equal to negative 10 the value is about 0.0000454 — tiny but positive. That is the same fact that gave every exponential function in Lessons 7.1 and 7.2 a range of the positive numbers and the horizontal axis as an asymptote.
Section
Section 3
Concept
A function of the form a times e to the rx is a natural base exponential function. With a positive, it grows when r is positive and decays when r is negative, because e itself is greater than 1.
natural base exponential function — A function of the form y equals a times e to the rx. With a positive, it is a growth function when r is positive and a decay function when r is negative.
\[ y = ae^{rx} \]
Such a function can always be rewritten in the base form of Lessons 7.1 and 7.2: a times e to the rx equals a times the quantity e to the r, all to the x, so the base is e to the r.
Figure (svg): The graphs of e to the x and e to the negative x, one growing and one decaying
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 493-493 — Natural Base Functions
Picture it
E to the x and e to the negative x.
Figure (svg): The graphs of e to the x and e to the negative x, one growing and one decaying
Both pass through the point where y is 1, and at x equal to 1 one reaches 2.718 while the other drops to 0.368 — which is one over e.
Worked example
Example 3a, with the alternative base form.
\[ \text{Graph } y = 3e^{0.25x} \text{ and state the domain and range.} \]
Classify it
Why: The coefficient 3 is positive and the exponent's coefficient 0.25 is positive, so this grows.
Find two anchor points
Why: At x equal to 0 the value is 3; at x equal to 1 it is 3 times e to the 0.25, about 3.85.
\[ (0, 3)\text{ and } (1, 3.85) \]
Draw the curve
Why: It rises to the right and flattens toward the horizontal axis on the left.
State the domain and range
Why: Every real input is allowed and every output is positive.
Figure (svg): The solution to Worked example graph a natural base growth function shown as a ladder of expressions, one row per algebraic move
\[ y = 3e^{0.25x}, \quad y > 0 \]
Verify: rewrite it in base form
Why: Three e to the 0.25x is 3 times the quantity e to the 0.25, all to the x, and e to the 0.25 is about 1.28. So the function is about 3 times 1.28 to the x — an ordinary growth function with base 1.28, which the textbook points out in an Another Way note. Both forms describe the same curve.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 493-493
Sorting
Read the sign of r, assuming a is positive.
Sort into buckets
Sort each natural base function.
The size of a plays no part at all. A coefficient of one third with r equal to 4 grows far faster than a coefficient of 100 with r equal to 0.01.
Worked example
Guided Practice 6 and 7.
\[ \text{Graph } y = 2e^{0.5x} \text{ and } f(x) = \tfrac{1}{2}e^{-x}+1. \]
First: classify and anchor
Why: Positive coefficient, positive r, so growth; the y-intercept is 2 and at x equal to 1 the value is about 3.30.
\[ (0, 2), (1, 3.30) \]
First: state the domain and range
Why: No translation, so the asymptote is the horizontal axis.
Second: classify
Why: Positive coefficient with negative r, so decay, and there is a vertical shift of 1.
Second: anchor and state
Why: At x equal to 0 the value is one half plus 1, which is 1.5; the asymptote is the line y equals 1.
Figure (svg): The solution to Worked example two more natural base graphs shown as a ladder of expressions, one row per algebraic move
\[ y > 0; \qquad y > 1 \]
Verify: check the second's behaviour far to the right
Why: As x grows, e to the negative x shrinks toward zero, so the function approaches 1 from above. At x equal to 5 the value is about 1.003. The constant 1 is the asymptote, exactly as k was in Lessons 7.1 and 7.2.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 494-494
Trap
\[ y = \tfrac{1}{3}e^{4x} \]
Call it decay because the coefficient is a small fraction
Why: The one third is taken as a shrinking factor.
\[ \text{exponential decay} \quad \text{(wrong)} \]
The values are one third, about 18.2, about 993 — growing extremely fast. The coefficient only sets the starting height.
\[ r = 4 > 0 \;\Longrightarrow\; \text{growth} \]
Read the sign of r for the direction and a for the y-intercept
Why: Two independent readings, exactly as base and coefficient were in Lesson 7.2.
\[ y = \tfrac{1}{3}e^{4x}: \; \text{starts low, climbs steeply} \]
Exercise 2 of the lesson asks about precisely this function. A small coefficient with a large positive r is one of the commonest shapes in real growth models.
Fill the middle
Example 3a.
Fill in the blanks
y = 3e^3.85: \; \text___ x=1, \; y = 3e^___ \approx 3(1.284) \approx ___
Why: E to the 0.25 is about 1.284, so the value is about 3.85. The two anchor points are always a and a times e to the r, which is the same rule as a and a times b in the base form.
Prediction
Commit before reasoning.
Predict first
The function 3 e to the 0.25x can be written as 3 times b to the x. What is b?
Correct: e to the 0.25, about 1.28.
\[ ae^{rx} = a(e^r)^x \;\Longrightarrow\; b = e^r \]
Why: The power of a power rule turns e to the 0.25x into the quantity e to the 0.25, all to the x, so the base is e to the 0.25, which is about 1.284. Every natural base function is an ordinary exponential function in disguise, and every ordinary one can be rewritten with base e — which is why one chapter covers both.
Comparison
Fill the blanks. Two notations, one family.
Comparison matrix
| Question | y = a b^x | y = a e^(rx) |
|---|---|---|
| Growth when | b > 1 | r > 0 |
| Decay when | 0 < b < 1 | r < 0 |
| The y-intercept | a | a |
| Converting | b = e^r | r is the exponent that gives b |
The last row's second entry is a question this chapter cannot yet answer: finding r from b needs a logarithm, which is Lesson 7.4.
Section
Section 4
Concept
Natural base functions translate exactly as the others do, with the asymptote moving to y equals k. Subtracting a decaying term from a constant produces a curve that rises steeply and then levels off, which is a common shape in growth models.
\[ l = 337 - 276e^{-0.178t} \]
As the decaying term shrinks toward zero, the whole expression climbs toward the constant. That constant is a horizontal asymptote and an upper limit the quantity never reaches.
Figure (svg): A tiger shark's length against its age, rising toward a limiting length of three hundred thirty-seven centimetres
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 493-494 — Examples 3b and 4
Picture it
Example 4: length against age, levelling off near 337 centimetres.
Figure (svg): A tiger shark's length against its age, rising toward a limiting length of three hundred thirty-seven centimetres
A newborn is 61 centimetres, a three-year-old about 175 and a five-year-old about 224. The curve flattens because each year adds less than the last.
Worked example
Example 3b.
\[ \text{Graph } y = e^{-0.75(x-2)}+1 \text{ and state the domain and range.} \]
Classify the untranslated version
Why: The coefficient is 1 and r is negative 0.75, so it decays.
Read h and k
Why: The exponent contains x minus 2, so h is 2; the constant outside is 1.
\[ h = 2, k = 1 \]
Shift the graph
Why: Right 2 and up 1, moving the point at height 1 to the point where x is 2 and y is 2.
\[ (2, 2) \]
State the asymptote, domain and range
Why: The asymptote moves up to the line y equals 1.
Figure (svg): The solution to Worked example translate a natural base function shown as a ladder of expressions, one row per algebraic move
\[ y > 1, \quad \text{asymptote } y = 1 \]
Verify: check a point either side
Why: At x equal to 0 the exponent is 1.5, so the value is e to the 1.5 plus 1, about 5.48. At x equal to 2 it is 1 plus 1, or 2. The curve is falling and flattening toward 1, which the asymptote predicts and the two values confirm.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 493-493
Fill the middle
Example 4, at t equal to 3.
Fill in the blanks
l = 337 - 276e^162 \approx 337 - 276(0.586) \approx 337 - ___
Why: Two hundred seventy-six times 0.586 is about 162, so the length is about 175 centimetres. The subtracted term is what the shark still has left to grow, and it shrinks by about 16 percent every year.
Worked example
Example 4 and Guided Practice 9.
\[ \text{With } l = 337-276e^{-0.178t}, \text{ find the length at } t = 3 \text{ and at } t = 5. \]
Evaluate the exponential part at 3
Why: Negative 0.178 times 3 is negative 0.534, and e to that is about 0.586.
\[ e ^{-0.534}\text{ about } 0.586 \]
Complete the calculation
Why: Two hundred seventy-six times 0.586 is about 162, and 337 minus 162 is about 175.
\[ \text{about } 175 \text{cm} \]
Evaluate at 5
Why: Negative 0.89 gives about 0.411, and 276 times that is about 113.
\[ e ^{-0.89}\text{ about } 0.411 \]
Complete
Why: Three hundred thirty-seven minus 113 is about 224.
\[ \text{about } 224 \text{cm} \]
Figure (svg): The solution to Worked example the tiger shark model shown as a ladder of expressions, one row per algebraic move
\[ 175 \text{ cm}; \qquad 224 \text{ cm} \]
Verify: check the newborn length and the limit
Why: At t equal to 0 the exponential is 1, so the length is 337 minus 276, which is 61 centimetres — a plausible newborn. As t grows the exponential shrinks toward zero and the length climbs toward 337 centimetres, which the model treats as the maximum adult size.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 494-494
Error analysis
A student uses the shark model far into the future.
Annotate
On: \( l(50) = 337 - 276e^{-8.9} \approx 337, \text{ so a } 100\text{-year-old shark is about } 674 \text{ cm} \)
Read which way the exponential term enters. Added, it drives unbounded growth; subtracted from a constant, it produces growth that levels off.
Sorting
Look at how the exponential term enters.
Sort into buckets
Sort each model by its shape.
The third shape is the one that simple exponential functions cannot produce, and it is why models of physical growth so often take that form.
Prediction
Commit before reasoning.
Predict first
In the shark model, what is the meaning of the number 337?
Correct: The maximum length, approached but never reached.
\[ e^{-0.178t} \to 0 \;\Longrightarrow\; l \to 337 \]
Why: As t grows the exponential term shrinks toward zero, so the length climbs toward 337 centimetres. The length at birth is 337 minus 276, which is 61 centimetres, and the growth rate is governed by the 0.178 in the exponent. Reading each constant's role separately is what makes a model like this interpretable.
Ranking
Using a natural base model.
Put in order
Why: Step one is what turns a formula into a description: knowing that 337 is a limit and 61 the birth length makes every later answer checkable against common sense. Step three has to come before step four, since the exponential is inside the subtraction.
Section
Section 5
Concept
As the number of compoundings per year grows without bound, the compound interest formula approaches a simpler one: the amount after t years is the principal times e to the rt.
continuously compounded interest — Interest compounded at every instant. The amount after t years is P times e to the rt, where P is the principal and r the annual rate as a decimal.
\[ A = Pe^{rt} \]
This answers the question Lesson 7.1 left open. Compounding more often always helps a little, but never more than the continuous case, which is the ceiling.
Figure (svg): The compound interest formula converging to the continuous formula as the number of compoundings grows
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 494-495 — Continuously Compounded Interest
Picture it
The two formulas, and the values they give.
Figure (svg): The compound interest formula converging to the continuous formula as the number of compoundings grows
Daily compounding at 6 percent gives 4247.30 dollars and continuous gives 4247.35 — a difference of five cents on 4000 dollars.
Worked example
Example 5.
\[ \text{Find the balance on } 4000 \text{ dollars at } 6\% \text{ compounded continuously for } 1 \text{ year.} \]
Choose the formula
Why: Continuous compounding uses A equals P times e to the rt.
\[ A = P e ^{r t} \]
Substitute
Why: P is 4000, r is 0.06 and t is 1.
\[ 4000 e ^{0.06} \]
Evaluate the exponential
Why: E to the 0.06 is about 1.0618.
\[ \text{about } 1.0618 \]
Complete and round
Why: Four thousand times 1.0618 is about 4247.35.
\[ 4247.35\text{ dollars} \]
Figure (svg): The solution to Worked example continuous compounding shown as a ladder of expressions, one row per algebraic move
\[ A = 4000e^{0.06} \approx 4247.35 \]
Verify: compare with periodic compounding
Why: Quarterly compounding at 6 percent gives 4245.68 dollars and daily gives 4247.30. Continuous gives 4247.35, the largest of the three but by very little. Compounding frequency matters far less than the rate itself, which is worth knowing before comparing two savings accounts.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 495-495
Fill the middle
Example 5.
Fill in the blanks
A = 4000e^1.0618 = 4000 \times ___ \approx 4247.35
Why: E to the 0.06 is about 1.0618, so continuous compounding at 6 percent multiplies the balance by about 1.0618 each year. That number is the effective annual growth factor, slightly above 1.06 because the interest compounds within the year.
Worked example
Guided Practice 10 and 11.
\[ \text{Find the balance on } 2500 \text{ dollars at } 5\% \text{ compounded continuously after } 2, 5 \text{ and } 7.5 \text{ years, and the interest earned.} \]
Two years
Why: The exponent is 0.10, and e to that is about 1.1052.
\[ 2762.93\text{ dollars} \]
Five years
Why: The exponent is 0.25, and e to that is about 1.2840.
\[ 3210.06\text{ dollars} \]
Seven and a half years
Why: The exponent is 0.375, and e to that is about 1.4550.
\[ 3637.49\text{ dollars} \]
Find the interest in each case
Why: Subtract the principal of 2500 from each balance.
\[ 262.93, 710.06, 1137.49 \]
Figure (svg): The solution to Worked example three time periods shown as a ladder of expressions, one row per algebraic move
\[ 2762.93; \; 3210.06; \; 3637.49 \]
Verify: check that the interest grows faster than the time
Why: From 2 to 5 years the time went up by a factor of 2.5 and the interest by a factor of 2.7; from 5 to 7.5 the time rose by half and the interest by 60 percent. Interest grows faster than proportionally because earlier interest is itself earning interest — which is the whole point of compounding.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 495-495
Trap
\[ A = P\left(1+\tfrac{r}{n}\right)^{nt} \text{ with } n = \infty \]
Substitute infinity for n
Why: The continuous case is treated as the periodic formula with a very large n.
\[ \text{no value: the expression is not defined at } n = \infty \]
Infinity is not a number to substitute. The formula has a limit as n grows, and that limit is a different formula.
\[ A = Pe^{rt} \]
Use the continuous formula, which is the limit
Why: The limiting behaviour was worked out once and recorded as its own formula.
\[ \left(1+\tfrac{r}{n}\right)^{n} \to e^r \]
Very large finite values of n give answers extremely close to the continuous one, so in practice the two agree to the nearest cent for any n above a few hundred.
Matching
Only one case needs e.
Match the pairs
Why: The first three are the same formula with different values of n, and the last is its limit. The balances they produce rise in that order, but the gaps shrink quickly — the whole distance from annual to continuous is smaller than the gap from a 6 percent rate to a 6.1 percent one.
Prediction
Commit before reasoning.
Predict first
At 6 percent on 4000 dollars for a year, how much more does continuous compounding earn than daily?
Correct: About five cents.
\[ 4247.30 \;\to\; 4247.35: \; \text{a gain of } 0.05 \]
Why: Daily gives 4247.30 dollars and continuous gives 4247.35. The gain from going from quarterly to daily was about 1.62 dollars, and from daily to continuous only five cents — each increase in frequency buys less than the last, which is exactly what convergence to a limit means. Advertising continuous compounding is worth almost nothing financially.
Comparison
Fill the blanks. One is the limit of the other.
Comparison matrix
| Question | Periodic | Continuous |
|---|---|---|
| Formula | P(1 + r/n)^(nt) | P e^(rt) |
| Inputs needed | P, r, n, t | P, r, t |
| At 6 percent for 1 year on 4000 | 4245.68 quarterly, 4247.30 daily | 4247.35 |
| Which is larger | smaller, for any finite n | larger, and it is the ceiling |
The continuous formula is simpler as well as larger, since it needs one fewer input — which is part of why it is the standard form in later mathematics.
Comparison
Fill the blanks. One family, three notations.
Comparison matrix
| Form | Growth when | Best suited to |
|---|---|---|
| y = a(1 + r)^t | r > 0 | a percent change per period |
| y = a b^x | b > 1 | a known factor per step |
| y = a e^(rx) | r > 0 | a continuous rate |
| All three | describe the same curves | converting between them needs logarithms |
Converting from base form to natural base form means finding the exponent that turns e into b, which is exactly what Lesson 7.4's logarithm does.
Pattern
One routine for expressions, one for graphs and models.
A model of the form constant minus a decaying term rises toward that constant. The constant is a limit, not a rate.
OpenStax Algebra and Trigonometry 2e, §6.1 Exponential Functions §6.1
Check
Simplifying. Add, do not multiply.
Check your understanding
Simplify (5e^(-3x))^2.
Answer: A
Why: The outer power applies to both the 5 and the power of e: 5^2 = 25 and -3x times 2 = -6x.
Check
Classifying. Read r, not a.
Check your understanding
Is f(x) = (1/3)e^(4x) growth or decay?
Answer: A
Why: The sign of r decides the direction; the coefficient only sets the y-intercept.
Check
Continuous compounding.
Check your understanding
Find the balance on 4000 dollars at 6 percent compounded continuously for 1 year.
Answer: A
Why: A = 4000e^(0.06) = 4000(1.0618), which is about 4247.35.
Real world
A cup of coffee at 90 degrees Celsius cools in a 20 degree room. Its temperature after t minutes is modelled by T equals 20 plus 70 times e to the negative 0.06t.
Discussion prompt
Find the temperature after 10 and 30 minutes, say what the two constants 20 and 70 mean, and explain why the coffee never reaches room temperature according to the model.
Hint: The exponential term is added to a constant here, not subtracted from one.
Answer:
\[ T(10) = 20 + 70e^{-0.6} \approx 20 + 38.4 = 58.4 \text{ degrees} \]
\[ T(30) = 20 + 70e^{-1.8} \approx 20 + 11.6 = 31.6 \text{ degrees} \]
After 10 minutes the coffee is about 58 degrees and after 30 minutes about 32 degrees. The 20 is the room temperature, the limit the coffee approaches, and the 70 is the initial excess over that room temperature, since 90 minus 20 is 70.
The model has the coffee approaching 20 degrees without reaching it, because e to a negative multiple of t shrinks toward zero but never equals it. That is the same asymptote behaviour as the shark model, with the exponential added to the constant rather than subtracted, so the curve falls toward the limit instead of rising to it. In reality the coffee does reach room temperature, which is a reminder that an asymptote is a feature of the model rather than of the world.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the function y equals one third times e to the 4x an example of growth or decay?
Correct: Growth — the exponent's coefficient is positive.
\[ \tfrac{1}{3}e^{4x}: \; \tfrac{1}{3}, \; 18.2, \; 993, \; 54{,}200 \]
Why: The values are one third, about 18.2, about 993 and about 54,200 at x equal to 0, 1, 2 and 3. The coefficient one third sets only the starting height; the direction comes entirely from the sign of r. Exercise 2 of the lesson asks about exactly this function, and mistaking a small coefficient for decay is the commonest error in the natural base form — the same mistake as reading two sevenths times 4 to the x as decay in Lesson 7.2.
Explain it
They have met e and think it is a variable.
Discussion prompt
In four sentences or fewer, explain what e is and why anyone bothered to name it.
Hint: Compare it with pi.
Answer:
The letter e stands for a specific number, about 2.718, in the same way that pi stands for about 3.142. It is what you get if you take one plus one over n and raise it to the power n, then let n get larger and larger — the values settle down rather than running away.
It was named because that limit turns up whenever something grows continuously rather than in steps: interest compounded every instant, a population growing all the time, a cup of coffee cooling. Every rule you know about exponents works on it, because it is just a number.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For simplifying, treat e as a number and name the property at each step. For direction, look only at the sign of r. For models, write down what each constant means before substituting anything. For interest, use e when the problem says continuously and the n-formula otherwise. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a one-page reference for e. Top left: write the limit definition and copy the table of six values, circling where the digits stop changing. Top right: simplify four expressions containing e that you invent, naming the property used at each step and leaving no negative exponents. Middle: sketch y equals e to the x and y equals e to the negative x on one set of axes, marking the shared point and the values at x equal to 1. Bottom left: take the shark model and evaluate it at t equal to 0, 3, 5 and 20, then sketch it and mark the asymptote, writing what each of 337, 276 and 0.178 means. Bottom right: compute the balance on 1000 dollars at 5 percent for 10 years three ways — annually, daily and continuously — and note the gaps between them. In a margin, write the identity converting a times e to the rx into base form.
If your three interest figures differ by more than a few dollars, recheck the periodic ones: the rate must be divided by n and the years multiplied by n.
Recap
Five things, and the first explains a constant that will appear in every later chapter.
| If you see | Then |
|---|---|
| e in an expression | Treat it as a number greater than 1 |
| a e^(rx) with r positive | Exponential growth |
| a e^(rx) with r negative | Exponential decay |
| A constant minus a decaying term | Growth levelling off toward the constant |
| A constant plus a decaying term | Decay levelling off toward the constant |
| The word continuously | Use A = P e^(rt) |
| A negative exponent in an answer | Rewrite it as a reciprocal |
Every question so far has run forwards: given the exponent, find the value. Lesson 7.4 runs it backwards, introducing the logarithm as the inverse of an exponential function.
McDougal Littell Algebra 2 (Texas Edition), Ch. 7 Exponential and Logarithmic Functions — Lesson 7.3 Use Functions Involving e §7.3, pp. 492-497 — everything on these slides traces back here
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