Linear versus exponential growth
One pay plan adds the same raise every day, another raises the pay by the same percent every day. Which one ends up paying more, and when?
Where you are 6 in Exponential growth and decay
Why this exists
A summer job offers two pay plans. Plan A pays 80 dollars on day 0, the first day, and adds a raise of 4 dollars every day after. Plan B pays only 40 dollars on day 0, then raises the pay by 10\% every day. For the first week Plan A is far ahead. Is it the better plan?
The question is not really about these numbers. It is about two kinds of change: adding the same amount each step, and multiplying by the same factor each step. Linear and nonlinear functions tells a line from a curve. This star puts the two most common kinds of change side by side, shows how to tell them apart from a table, and asks which one wins in the long run.
Definition
Equal intervals of input are steps of the same length h: from x to x + h, from x + h to x + 2h, and so on.
A function grows linearly when, over equal intervals, it changes by equal differences: f(x + h) - f(x) is the same number wherever the interval starts. Linear functions f(x) = mx + c behave this way.
A function grows exponentially when, over equal intervals, it changes by equal factors: \dfrac{g(x + h)}{g(x)} is the same number wherever the interval starts. Exponential functions g(x) = a \cdot b^x, with a \neq 0, b > 0 and b \neq 1, behave this way, and b is the factor for one unit (Growth and decay factors).
In a table with equally spaced inputs, subtracting each output from the next tests for equal differences, and dividing each output by the one before tests for equal factors. Differences of outputs divided by differences of inputs are the average rates of change already met; for a linear function they all equal the slope.
Intuition
Here are the two pay plans for the first five days, before any rounding to the cent.
| day x | Plan A (dollars) | Plan B (dollars) |
|---|---|---|
| 0 | 80 | 40 |
| 1 | 84 | 44 |
| 2 | 88 | 48.4 |
| 3 | 92 | 53.24 |
| 4 | 96 | 58.564 |
Plan A’s pay goes up by 4 every day. Plan B’s goes up by 4, then 4.4, then 4.84, then 5.324: each raise is 10\% of a bigger pay, so the raises themselves grow. What stays fixed for Plan B is the factor, 1.1 a day.
A fixed step against a step that keeps growing: once the growing step is the bigger one, the exponential closes the gap a little faster every day, and in the end it gets past any head start. Figure 1 shows the same race with a much bigger head start, the line y = 500 + 100x against y = 2^x.
For most of the picture the curve hugs the axis while the line is far above it. Then, within a couple of steps, the curve shoots past.
Results
Equal differences. For f(x) = mx + c and any interval length h,
Equal factors. For g(x) = a \cdot b^x and any interval length h,
Reading a table. Outputs at equally spaced inputs with equal differences fit a linear function; with equal ratios they fit an exponential function. Outputs with neither fit neither.
An increasing exponential passes every increasing line. If a > 0 and b > 1, then however large c and m are, a \cdot b^x > mx + c for every large enough x. Before that the line may lead for a long time, so which is bigger depends on where you look.
A falling line against decay. If m < 0 and c > 0, the line mx + c reaches 0 and is negative after that. If a > 0 and 0 < b < 1, then a \cdot b^x > 0 for every x: the decay comes ever closer to 0 but never reaches it, so it ends up above the falling line.
Why it is true
Linear. Expand and subtract:
No x is left, so the change is the same wherever the interval starts.
Exponential. By Multiplying and dividing powers with the same base, b^{x + h} = b^x \cdot b^h, so
Again no x is left. Dividing is allowed because a \cdot b^x is never 0.
Why the exponential gets past the line. One step of g, from x to x + 1, adds a \cdot b^{x+1} - a \cdot b^x, which is a \cdot b^x(b - 1), and the next step is b times this one, because moving x on by 1 multiplies a \cdot b^x by b. With b > 1 the steps keep growing, so they eventually exceed the line’s fixed step m, and from then on g gains on the line by more each step than the step before. The tables and graphs below show where the crossing lands; a full proof comes in Linear, polynomial and exponential growth compared.
Decay stays positive. With a > 0, a \cdot b^x is a product of positive numbers, because a positive base to any power is positive.
Worked examples
Four tables
Say whether each table shows equal differences, equal factors or neither.
| x | table A | table B | table D |
|---|---|---|---|
| 0 | 3 | 3 | 1 |
| 1 | 7 | 6 | 2 |
| 2 | 11 | 12 | 4 |
| 3 | 15 | 24 | 7 |
| 4 | 19 | 48 | 11 |
Table C has inputs x = 0, 2, 4, 6, 8 and outputs 5, 15, 45, 135, 405.
SolutionThe inputs in A, B and D go up by 1 each time, so the intervals are equal and the outputs can be compared directly.
Table A: the differences are 7 - 3 = 4, then 4, 4, 4. Equal differences, so A grows linearly, by 4 per unit.
Table B: the differences 3, 6, 12, 24 are not equal, but the ratios are: \tfrac{6}{3} = 2, \tfrac{12}{6} = 2, and so on. Equal factors, so B grows exponentially, doubling per unit.
Table C: the inputs go up by 2, still equal intervals. The ratios are \tfrac{15}{5} = 3, \tfrac{45}{15} = 3, \tfrac{135}{45} = 3, \tfrac{405}{135} = 3: a factor of 3 every 2 units, so C grows exponentially.
Table D: the differences are 1, 2, 3, 4 and the ratios are 2, 2, 1.75, about 1.57. Neither is constant, so D is neither linear nor exponential. Its differences grow by the same amount each time, a pattern of its own.
When does Plan B pay more?
Plan A pays 80 + 4x dollars on day x and Plan B pays 40(1.1)^x dollars. On which day does Plan B first pay more?
SolutionPlan B’s daily raise is 10\% of its pay: 4 dollars on day 1, the same as Plan A’s, and more than 4 dollars on every day after that, so from day 2 on Plan B gains on Plan A every day. But it starts 40 dollars behind, so the catching up takes a while. Multiplying by 1.1 day after day gives 40(1.1)^{10} \approx 103.75 dollars on day 10, still behind Plan A’s 120. The next days:
| day x | 11 | 12 | 13 |
|---|---|---|---|
| Plan A (dollars) | 124 | 128 | 132 |
| Plan B (dollars) | 114.12 | 125.54 | 138.09 |
On day 12 Plan A still pays more; on day 13 Plan B does. Plan B first pays more on day 13, and after that its lead grows every day. For a job of a week or two, Plan A pays more; for a longer one, Plan B eventually does.
A huge head start for the line
Compare f(x) = 500 + 100x and g(x) = 2^x for whole numbers x \geq 0. Which is bigger, and where?
SolutionThe line starts 499 ahead and climbs 100 a step; the exponential starts at 1 and doubles. A table with a few chosen rows:
| x | 500 + 100x | 2^x |
|---|---|---|
| 0 | 500 | 1 |
| 5 | 1{,}000 | 32 |
| 9 | 1{,}400 | 512 |
| 10 | 1{,}500 | 1{,}024 |
| 11 | 1{,}600 | 2{,}048 |
| 12 | 1{,}700 | 4{,}096 |
For x = 0 to 10 the line is bigger; at x = 9 it is still nearly three times as big. At x = 11 the exponential is bigger, and at x = 12 it is already more than twice the line. The graphs cross between x = 10 and x = 11, as Figure 1 shows, and from x = 11 on 2^x stays ahead: each step adds 2^x to it, which is already more than the line’s 100.
So “which is bigger?” has two answers. Up to x = 10, the line. From x = 11 on, the exponential, by more every step.
Faster is not the same as bigger
A band’s fan page has 6{,}000 followers and gains 300 a week. A second band’s page has 2{,}000 followers and grows 8\% a week. After w weeks they have F(w) = 6000 + 300w and G(w) = 2000(1.08)^w followers. When does the second page first gain more in a week than the first, and when does it first have more followers?
SolutionThe second page’s gain in a week is 8\% of its followers at the start of that week. It gains more than 300 once G(w) > 3750, since 0.08 \times 3750 = 300. G(8) \approx 3701.86 and G(9) \approx 3998.01, so the week from 8 to 9 adds about 296.15, and the week from 9 to 10 adds about 319.84. From week 9 to 10 on, the second page grows faster.
Being bigger takes far longer, because the gap built up meanwhile must be closed:
| week w | F(w) | G(w), rounded |
|---|---|---|
| 10 | 9{,}000 | 4{,}318 |
| 20 | 12{,}000 | 9{,}322 |
| 24 | 13{,}200 | 12{,}682 |
| 25 | 13{,}500 | 13{,}697 |
| 30 | 15{,}000 | 20{,}125 |
The second page first has more followers after 25 weeks, sixteen weeks after it started growing faster. By week 30 it is ahead by about 5{,}125.
A million-follower head start
Show that 1.5^x passes 1{,}000{,}000 + 1{,}000x, and find the first whole x where it is bigger.
SolutionTen steps of 1.5 multiply by 1.5^{10} \approx 57.67, so jumping ten at a time: 1.5^{10} \approx 57.67, 1.5^{20} \approx 3{,}325, 1.5^{30} \approx 191{,}751. At x = 30 the line is 1{,}030{,}000, still more than five times as big. Four more steps multiply by 1.5^4 = 5.0625:
| x | 34 | 35 |
|---|---|---|
| 1{,}000{,}000 + 1{,}000x | 1{,}034{,}000 | 1{,}035{,}000 |
| 1.5^x, rounded | 970{,}740 | 1{,}456{,}110 |
The first whole x where 1.5^x is bigger is 35, and from there it stays ahead: each further step adds half of 1.5^x, already more than 700{,}000, against the line’s 1{,}000. A head start of a million cost the exponential only 35 steps, because each step multiplies its value while the line adds a fixed 1{,}000.
Decay against a falling line
Compare f(x) = 100 - 10x and g(x) = 100(0.8)^x for x \geq 0.
SolutionBoth start at 100. The line loses 10 every step. The decay loses 20\% of what is left: 20, then 16, then 12.8, steps that shrink.
| x | f(x) | g(x), rounded |
|---|---|---|
| 0 | 100 | 100 |
| 2 | 80 | 64 |
| 4 | 60 | 40.96 |
| 6 | 40 | 26.21 |
| 8 | 20 | 16.78 |
| 9 | 10 | 13.42 |
| 10 | 0 | 10.74 |
| 12 | -20 | 6.87 |
At first the decay falls faster, so the line is above it, for x = 1 to 8. The decay’s steps keep shrinking while the line’s stay 10, and between x = 8 and x = 9 the graphs cross. The line reaches 0 at x = 10 and is negative after, while g(x) stays positive for every x, coming ever closer to 0. Figure 2 shows the two; the x-axis is an asymptote of the decay curve.
Boundaries
The intervals must be equal. Differences over steps of different lengths cannot be compared directly: a jump of 6 over 2 units and a jump of 3 over 1 unit are the same rate. Divide by the length of each step, or use equal steps.
A table shows only its rows. Equal differences or equal ratios in a table say the rows fit a linear or exponential function, not what happens between or beyond them.
A zero output rules out equal factors. a \cdot b^x is never 0, and a ratio with 0 underneath is undefined.
“Eventually passes” needs a > 0 and b > 1. With 0 < b < 1 the exponential shrinks, and an increasing line stays above it from some point on. The crossing may also be far out, as at x = 35 in Example 5.
Crossings here are found to the whole number. Tables and graphs place a crossing between two whole numbers; a finer table or a closer look at the graph (Solving f(x) = g(x) with graphs and tables) narrows it further.
Where this goes next
Graphs of exponential functions draws these curves in detail, and Writing an exponential function from a table or graph writes the rule once the equal factor is found. Choosing a function family for data uses the equal-differences and equal-factors tests to decide which model fits a situation. Listed as sequences, equal differences are an arithmetic sequence (Arithmetic sequences and the explicit formula) and equal factors a geometric one (Geometric sequences and the explicit formula). Table D in Example 1 has differences that grow by equal amounts, the mark of a quadratic; Linear, polynomial and exponential growth compared compares linear, polynomial and exponential growth and proves that the exponential wins.
Exercises
Equal differences or equal factors
Say whether each table shows equal differences, equal factors or neither, and give the difference or factor.
- x: 0, 1, 2, 3, 4; y: 7, 10, 13, 16, 19.
- x: 0, 1, 2, 3, 4; y: 5, 10, 20, 40, 80.
- x: 0, 1, 2, 3, 4; y: 400, 200, 100, 50, 25.
- x: 0, 1, 2, 3, 4; y: 2, 3, 5, 8, 12.
- x: 0, 3, 6, 9; y: 10, 40, 160, 640.
- x: 0, 1, 3, 4; y: 8, 11, 17, 20.
Any interval
- For f(x) = 12 + 5x, find f(x + 4) - f(x), and check it with f(2) and f(6).
- For g(x) = 3(2)^x, find \dfrac{g(x + 3)}{g(x)}, and check it with g(1) and g(4).
- For h(x) = 1000(0.9)^x, find the factor over any 2 units, and the percent lost over 2 units.
Pay plans and followers
- Plan A pays 70 + 5x dollars on day x and Plan B pays 35(1.12)^x dollars. Which pays more on day 5, and on day 10? On which day does Plan B first pay more?
- A band’s page has 9{,}000 followers and gains 300 a week; another has 1{,}500 and grows 8\% a week. After how many whole weeks does the second page first have more followers?
- Two plans both pay 60 dollars on day 0. Plan A adds 6 dollars a day and Plan B raises the pay 10\% a day. Compare them on days 1, 2, 3 and 4, and explain why Plan B pays more on every day after day 1.
Which is ahead, and where
- Find the first whole x where 2^x is bigger than 1000 + 100x.
- Compare f(x) = 50 + 20x and g(x) = 1.5^x at x = 5, 10 and 15, and find the first whole x where g is bigger.
Decay and a falling line
- For f(x) = 64 - 8x and g(x) = 64(0.75)^x, compare f(4) and g(4), find where f reaches 0 and the value of g there, and compare f(10) and g(10). Does g ever reach 0?
- Find the first whole x where f(x) = 120 - 15x is smaller than g(x) = 120(0.5)^x.
Reasoning
- Gorka says 2^x can never catch 2000 + 500x, because the line starts 1{,}999 ahead and climbs 500 a step. Is he right? Find the first whole x where 2^x is bigger.
- For g(x) = a \cdot b^x with a > 0 and b > 1, show that the step g(x + 1) - g(x) equals a \cdot b^x(b - 1) and that each step is b times the one before. Explain why this means g’s steps eventually beat any fixed slope m, however large.
Answers
- Equal differences of 3: 10 - 7 = 13 - 10 = 16 - 13 = 19 - 16 = 3. Linear.
- Equal factors of 2: each output is double the one before. Exponential.
- Equal factors of \tfrac{1}{2}: each output is half the one before. Exponential decay.
- Neither. The differences are 1, 2, 3, 4 and the ratios 1.5, about 1.67, 1.6, 1.5.
- Equal factors of 4 every 3 units: \tfrac{40}{10} = \tfrac{160}{40} = \tfrac{640}{160} = 4. Exponential.
- Equal differences of 3 per unit. The steps are not equal: 11 - 8 = 3 over 1 unit, 17 - 11 = 6 over 2 units, 20 - 17 = 3 over 1 unit, and 6 \div 2 = 3. Linear.
- f(x + 4) - f(x) = 5 \times 4 = 20. Check: f(2) = 22 and f(6) = 42, and 42 - 22 = 20.
- \dfrac{g(x + 3)}{g(x)} = 2^3 = 8. Check: g(1) = 6 and g(4) = 48, and 48 \div 6 = 8.
- The factor is 0.9^2 = 0.81, so 1 - 0.81 = 0.19: 19\% is lost over any 2 units.
- Day 5: Plan A 95 dollars, Plan B about 61.68. Day 10: Plan A 120, Plan B about 108.70. Day 11: 125 against about 121.75; day 12: 130 against about 136.36. Plan B first pays more on day 12.
- After 33 weeks. At 32 weeks: 9000 + 300 \times 32 = 18{,}600 against 1500(1.08)^{32} \approx 17{,}606. At 33 weeks: 18{,}900 against about 19{,}014.
- Day 1: 66 and 66, equal. Day 2: 72 against 72.6. Day 3: 78 against 79.86. Day 4: 84 against about 87.85. From day 1 on, Plan B’s pay is at least 66, so its daily raise is at least 6.6 dollars, more than Plan A’s 6: it pulls further ahead every day.
- x = 12. At x = 11: 2^{11} = 2048, less than 1000 + 1100 = 2100. At x = 12: 2^{12} = 4096, more than 2200.
- At 5: 150 against about 7.59. At 10: 250 against about 57.67. At 15: 350 against about 437.89. At 14: 330 against about 291.93, so the first whole x is 15.
- f(4) = 32 and g(4) = 20.25. f(8) = 0, and g(8) \approx 6.41. f(10) = -16 and g(10) \approx 3.60. No: g(x) is 64 times a positive power, so it is positive for every x.
- x = 8. At x = 7: f(7) = 15 against g(7) = 0.9375. At x = 8: f(8) = 0 against g(8) = 0.46875.
- No. At x = 13: 2^{13} = 8192, less than 2000 + 6500 = 8500. At x = 14: 2^{14} = 16384, more than 9000. From there each step adds 2^x, far more than 500.
- The step is g(x + 1) - g(x), which is a \cdot b^x \cdot b - a \cdot b^x, and taking out the common factor gives a \cdot b^x(b - 1). The next step is a \cdot b^{x+1}(b - 1), which is b times the first. The steps start at the positive number a(b - 1) and are multiplied by b > 1 again and again, so they grow without limit, like a(b - 1) \cdot b^x itself, and sooner or later pass any fixed m.