1
The base carries the whole story
a is the value at x = 0 and nothing more; it stretches the curve vertically. Growth against decay, and how fast, are questions about b alone.
Lesson · Exponential and Logarithmic Functions · Exponential functions
Growth and Decay
For a positive starting quantity, the same formula a·b^x describes an amount that keeps multiplying up and one that keeps shrinking. Which of the two you have is decided by the base, and the percentage in the problem statement is that base in disguise.
f(x) = a·b^x with a > 0
b > 1 growth up by (b − 1)·100% each step
0 < b < 1 decay down by (1 − b)·100% each step
b = 1 neither the constant a, for ever
One formula, three regimes
Core ideas
What this lesson establishes
The base does all the work. The starting value scales the picture and changes nothing about its shape.
2
A percentage change is a multiplier
Up 5% means multiply by 21/20. Down 20% means multiply by 4/5. Writing b = 1 + r for a rise and b = 1 − r for a fall is the step that turns a sentence about percentages into a function.
3
Constant percentage, changing amounts
A quantity growing 8% a year gains more each year than it did the year before, because 8% is taken of a larger number every time. The percentage is what stays fixed; the amount is what does not, and confusing the two is the error the worked example is about.
The material
Growth, decay, and how fast
Reading the base, writing a percentage as a multiplier, and the two different things people mean by how fast.
Definition · Growth factor
In f(x) = a·b^x with a > 0, the base b is the growth factor. If b > 1 the function increases and the model is one of growth; if 0 < b < 1 it decreases and the model is one of decay. b = 1 gives the constant a, which is neither.
One step multiplies by b, so the change across one step is f(x + 1) − f(x) = a·b^x·(b − 1) = (b − 1)·f(x). Divide by f(x): the change is the fraction b − 1 of the current value, the same fraction at every x. Written as a percentage, that fraction is the rate.
up 5% b = 1 + 5/100 = 21/20
up 1% b = 1 + 1/100 = 101/100
up 100% b = 1 + 1 = 2 doubling each step
down 20% b = 1 − 20/100 = 4/5
down 10% b = 1 − 10/100 = 9/10
down 100% b = 0 not an exponential
Keep the base as a fraction. 21/20 raised to a power is exact; 1.05 raised to a power on a machine is not, and the difference shows in the fourth or fifth place after ten steps.
Doubling time and half-life
Definition · Doubling time and half-life
The doubling time of a growth model is the number of steps n with b^n = 2. The half-life of a decay model is the n with b^n = 1/2. Both depend on b alone.
f(x + n) = a·b^(x+n) = a·b^x·b^n = f(x)·b^n. So f(x + n) is 2·f(x) exactly when b^n = 2, whatever x is and whatever a is. A quantity therefore takes exactly as long to go from 1000 to 2000 as from 1 to 2, and the doubling time is a property of the base rather than of where you started.
□
b^n = 2 rarely has a whole-number solution. For b = 101/100 the powers step across 2 between n = 69 and n = 70: (101/100)^69 ≈ 1.9869 and (101/100)^70 ≈ 2.0068, both rounded to four places. The exact answer is irrational, near 69.6607. Solving b^n = 2 for n is what “Solving Exponential Equations” does, and it needs a logarithm.
Example · The rule of 70
For a rise of r percent a step, the doubling time is close to 70/r. At 1% that gives 70 against a true 69.6607; at 5% it gives 14 against 14.2067; at 10% it gives 7 against 7.2725. For now this is a checked shortcut, not a derived rule: “Solving Exponential Equations” will solve the doubling equation after logarithms have been defined. The rule runs slightly high below about 2% a step and slightly low above it, and by 20% a step it offers 3.5 against a true 3.8018.
Decay is growth reflected
(1/b)^x = b^(−x), so the decay curve with base 1/b is the growth curve with base b reflected in the vertical axis. There is one shape here seen twice, not two shapes. It also means decay never reaches zero: b^n is a positive fraction for every n, however large, so a substance losing 10% a year still has (9/10)^100 of itself left after a century.
This is the point where an exponential model and a linear one give genuinely different answers, not just different numbers. Something losing 10 g a year from 100 g is empty in ten years. Something losing 10% a year never is.
Interactive
Growth and decay
The base decides which, and exactly how fast
Method
Setting up a growth or decay model
The modelling is four decisions. None of them is arithmetic.
1
Name the starting amount
a = f(0): the value before any step has been taken. If the data begins at year 3, either shift the input or say plainly that x counts steps from there.
2
Turn the percentage into a base
Up r percent gives b = 1 + r/100; down r percent gives b = 1 − r/100. Write it as a fraction and the powers stay exact.
3
Check the direction before computing anything
b > 1 has to go with a quantity you expect to grow. If you wrote b = 0.05 for a 5% rise, the model loses 95% a step and the first two values will show it.
4
Say what one step is
b is the multiplier per step, and a step is a year, an hour or a half-life — whatever the problem says. Nothing in a·b^x records the unit, so the answer has to.
Base, direction, amount
Score 0 / 0Choose the answer you can justify.
500 g losing 20% a year
The base first, then the table, then the half-life. The last column is the one the lesson is about.
starting amount a = 500 g
down 20% a year b = 1 − 20/100 = 4/5
A(n) = 500·(4/5)^n
n A(n) exactly lost during year n
──────────────────────────────────────────────────
0 500
1 400 100
2 320 80
3 256 64
4 1024/5 = 204.8 51.2
5 4096/25 = 163.84 40.96
half-life: (4/5)³ = 64/125 = 0.512 still above half
(4/5)⁴ = 256/625 = 0.4096 below half
so the first whole year at or under half is n = 4,
and the exact half-life is irrational, ≈ 3.1063 years
The right-hand column is the point. The percentage lost is 20 in every single year; the amount lost falls from 100 g to 40.96 g across five years, because 20% of a smaller number is a smaller number.
A reader who models this as "loses 100 g a year" gets year one right and year five badly wrong: that model is at 0 g after five years, and the true figure is 163.84 g. The two answers are not close, and they were built from the same sentence.
For a faded model, take 240 units growing by 12% each hour. The first move is supplied: b = 1 + 12/100 = 28/25. Write the function with its time unit, find the amount after two hours without rounding the base, and explain why the second hourly gain differs from the first. Check that consecutive values have ratio 28/25.
Common mistakes
1
Taking the percentage of the original each timeDown 20% from 500 is 400; down 20% again is 320, not 300. The second reduction is taken from 400. Repeated percentage change is multiplication, and it multiplies the current value.
2
Using the rate as the baseA 5% rise is b = 1.05, not b = 0.05 and not b = 5. The 1 is the amount you keep and the 0.05 is what is added to it; dropping the 1 converts a small rise into a near-total collapse.
3
Mixing the unit of the base with the unit of the exponentIf b = 21/20 is per year, then x = 18 means eighteen years. Going monthly is not dividing 5% by 12: it is finding the c with c¹² = 21/20, which is a different number. “Compound Interest and Continuous Growth” gives that distinction its own name.
Check your understanding
✓
Finish when a sentence turns into a base without hesitation.
Given "falls 12% an hour from 90 mg" you should write A(t) = 90·(22/25)^t with t in hours, say that it decays, and say that it never reaches 0. If the base takes you more than a line, the percentage-to-multiplier step is what to drill.
Note: Every model here changes by a fixed factor per whole step. “The Number e” asks what happens when the step is cut smaller and smaller instead — the same 100% rise, applied twice a year, then monthly, then daily — and the answer to that question is a specific number.