Exponential models
Bank balances, town populations, the value of a used phone: some quantities change by the same percent every step, not by the same amount. Each step multiplies by the same number. Questions about these exponential models appear on almost every SAT, and they reward careful reading more than heavy calculation.
By the end of this lesson you can
- name the three parts of an exponential model and read each one from an equation or a graph;
- turn a percent change, or words like "doubles", into a growth factor;
- tell a linear pattern from an exponential one;
- avoid the traps the SAT sets in exponential questions.
The three parts of a model
Exponential model
$$y = a(b)^x$$- \(a\) is the starting value: the value of \(y\) when \(x = 0\). On a graph it is the \(y\)-intercept.
- \(b\) is the growth factor: the multiplier applied once per step.
- \(x\) counts the steps: years, hours, months, or whatever the question uses.
If \(b > 1\), the quantity grows. If \(0 < b < 1\), it decays. Most SAT exponential questions become straightforward once you can point at \(a\), \(b\) and \(x\) and say what each one means in the story.
Explore \(y = a(b)^x\)
Interactive
Try this
- Change \(a\) only. Which point on the graph moves, and which features stay the same?
- Set \(b\) to 0.5. What happens to the curve as \(x\) grows? Does it ever reach zero?
- Find a value of \(b\) that makes \(y\) grow by 25% per step.
From percent to factor
Most SAT questions describe the change in a sentence, so step one is translating that sentence into \(a\) and \(b\). Remember: a percent change is not the factor. Turn it into a decimal, then combine it with 1: plus for growth, minus for decay.
Percent change to growth factor
- Increases by \(r\%\): \(b = 1 + \dfrac{r}{100}\)
- Decreases by \(r\%\): \(b = 1 - \dfrac{r}{100}\)
- Doubles: \(b = 2\). Triples: \(b = 3\). Is cut in half: \(b = \tfrac12\).
| The question says... | Growth factor \(b\) |
|---|---|
| rises by 15% each year | 1.15 |
| grows by 4% each month | 1.04 |
| drops by 30% each hour | 0.70 |
| keeps 92% of its value each year | 0.92 |
| triples every step | 3 |
| loses half its mass every step | 0.5 |
Quick check
A town's population decreases by 6% each year. What is the growth factor?
- 0.06
- 1.06
- 0.94
- 6
Show solution
Losing 6% means keeping 94%: \(b = 1 - 0.06 = 0.94\).
Linear or exponential?
In a linear pattern, the same number is added every step. In an exponential pattern, every step multiplies by the same number. If a question shows a table, decide which kind it is before writing an equation.
| \(x\) | 0 | 1 | 2 | 3 | Pattern |
|---|---|---|---|---|---|
| Linear | 4 | 7 | 10 | 13 | adds 3 each time: \(y = 3x + 4\) |
| Exponential | 4 | 8 | 16 | 32 | doubles each time: \(y = 4(2)^x\) |
Test for exponential: divide each value by the one before it. If you always get the same number, that number is \(b\).
Worked example 1
At \(x = 0\), \(y = 400\). Every time \(x\) increases by 1, \(y\) falls by 15%. Write \(y\) as a function of \(x\).
- Starting value. The value at \(x = 0\) is \(a = 400\).
- Growth factor. A 15% decrease keeps 85%: \(b = 1 - 0.15 = 0.85\).
- Put them into the form. \(y = 400(0.85)^x\)
- Check one step. At \(x = 1\), \(y = 400(0.85) = 340\), which is 15% less than 400.
\(y = 400(0.85)^x\)
When the exponent is not just \(x\)
Sometimes one "step" is not one unit of the input. A quantity that doubles every 3 hours does not double every hour. Its model is
$$N = a(2)^{t/3}$$because after \(t\) hours it has gone through \(t/3\) doublings. Watch for exponents like \(2t\), \(t/4\) or \(12t\). To find the change for one unit of the input, rewrite the exponent with the power rule: for example, \((1.1)^{2t} = \left((1.1)^2\right)^t = (1.21)^t\).
Worked example 2
A sample of bacteria triples every 5 hours. After 15 hours there are 8,100 bacteria. How many were there at the start?
- Count the steps. One step is 5 hours, so 15 hours is \(15 \div 5 = 3\) steps.
- Write the model. \(N = a(3)^{t/5}\), with \(a\) unknown.
- Substitute. \(8100 = a(3)^3 = 27a\)
- Solve. \(a = 8100 \div 27 = 300\)
300 bacteria. (Dividing 8,100 by 3 just once, to get 2,700, is the trap.)
Practice in the style of the SAT
Work out your answer first, then open the solution. Every question hides one of the mistakes the test likes to reward.
Question 1 · Interpreting a model
\(V = 24{,}000(0.92)^t\)
The equation estimates the value \(V\), in dollars, of a delivery van \(t\) years after its purchase. Which of the following is the best interpretation of 0.92 in this context?
- The van loses 92% of its value each year.
- Each year, the van is worth 92% of its value from the year before.
- The van was bought for 92% of 24,000 dollars.
- The van's value decreases by 0.92 dollars each year.
Show solution
0.92 is the factor you multiply by each year, so the van keeps 92% of its value and loses \(1 - 0.92 = 8\%\) per year.
The trap: choice A confuses the part kept with the part lost. Choice C describes the starting value, which is the full 24,000. Choice D treats the change as linear.
Question 2 · Hidden exponent
\(h(t) = 50(1.1)^{2t}\)
Each time \(t\) goes up by 1, the value of \(h(t)\) goes up by \(p\%\). What is \(p\)?
- 10
- 20
- 21
- 121
Show solution
Rewrite the exponent: \((1.1)^{2t} = \left((1.1)^2\right)^t = (1.21)^t\). For each increase of 1 in \(t\), \(h\) is multiplied by 1.21, an increase of 21%.
The trap: choice A ignores the 2 in the exponent. Choice B doubles 10% instead of compounding it. Choice D gives the factor as a percent, not the increase.
Question 3 · Student-produced response
The number of members of an online club doubles every 4 months. Twelve months after the club started, it had 9,600 members. How many members did it have when it started?
Show solution
12 months is \(12 \div 4 = 3\) doublings, so \(9600 = a(2)^3 = 8a\) and \(a = 1200\).
The trap: dividing by 2 only once (4,800) or dividing by 3 (3,200).
Question 4 · Building a model
A lake contains 200 kilograms of a pollutant. A cleanup removes 25% of the pollutant remaining each week. Which equation gives the amount \(P\), in kilograms, remaining after \(w\) weeks?
- \(P = 200(0.25)^w\)
- \(P = 200 - 25w\)
- \(P = 200(1.25)^w\)
- \(P = 200(0.75)^w\)
Show solution
Removing 25% each week keeps 75%, so \(b = 0.75\) and \(a = 200\).
The trap: choice A uses the part removed as the factor. Choice B is linear: it removes 25 kg, not 25%. Choice C is growth.
Question 5 · From a table
| \(x\) | 0 | 1 | 2 |
|---|---|---|---|
| \(y\) | 6 | 9 | 13.5 |
The table shows values of an exponential function. What is the value of \(y\) when \(x = 4\)?
Show solution
Ratios: \(9 \div 6 = 1.5\) and \(13.5 \div 9 = 1.5\), so \(y = 6(1.5)^x\). Then \(y = 6(1.5)^4 = 6(5.0625) = 30.375\).
Before you calculate, ask three questions
- What is the start? That is \(a\), the value when the input is 0.
- What is the multiplier? That is \(b\). Get it from the percent change, or from words such as "doubles" or "halves".
- How many steps have passed? Check how the exponent is written. If it is \(2t\) or \(t/4\), a step is not the same as one unit of time.
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