Linear & Exponential Growth - PSAT Math

SECTION 1

Historical Context & Motivation

Humans have been tracking growth patterns for thousands of years — from ancient farmers measuring steady crop yields to bankers computing compound interest in Renaissance Italy. The distinction between linear growth (adding the same amount each period) and exponential growth (multiplying by the same factor each period) sits at the heart of algebra, finance, biology, and data analysis. Understanding which model fits a situation helps you make accurate predictions and avoid costly errors.

~300 BCE • Euclid's Arithmetic Progressions

Euclid described sequences that increase by a constant difference, laying the groundwork for what we now call linear sequences.

1494 • Pacioli & Compound Interest

Luca Pacioli published tables for compound interest in his accounting textbook, giving merchants an early tool for exponential calculations.

1798 • Malthus on Population

Thomas Malthus argued that population grows exponentially while food supply grows linearly, framing one of history's most famous growth-model debates.

1965 • Moore's Law

Gordon Moore observed that the number of transistors on a chip roughly doubles every two years — a modern icon of exponential growth in technology.

The central question this lesson addresses is straightforward: How do you tell whether a quantity is growing linearly or exponentially, and how do you write the equation for each? On the PSAT, you will see tables, graphs, and word problems that test exactly this skill, so building a solid mental model now will pay off on test day.

SECTION 2

Core Principles & Definitions

Before diving into formulas, you need to lock down three foundational ideas that separate linear behavior from exponential behavior. Every PSAT question on this topic relies on at least one of these principles, so treat them as your diagnostic checklist.

Constant Difference vs. Constant Ratio

In a linear pattern, consecutive outputs differ by the same amount (the slope). In an exponential pattern, consecutive outputs share the same ratio (the growth factor).

Rate of Change

Linear functions have a constant rate of change — the slope never varies. Exponential functions have a rate of change that itself increases (or decreases), which is why they curve on a graph.

Starting Value & Parameter

Both models begin from an initial value. For linear, the key parameter is the slope (m). For exponential, it is the base (b).

Long-Run Behavior

A linear function eventually reaches any target value in a predictable, steady fashion. An exponential growth function will outpace any linear function given enough time, no matter how steep the line.

SECTION 3

Visual Explanation — Linear vs. Exponential on a Graph

The most immediate way to distinguish the two growth types is to look at their graphs. A linear function produces a straight line, while an exponential function produces a curve that starts gently and then sweeps upward (or downward for decay). The diagram below plots both side by side using the same starting value so you can see exactly where they diverge.

The blue line represents linear growth (y = 10 + 10x), increasing by the same amount each step. The violet curve represents exponential growth (y = 10 × 1.5ˣ), which starts slowly but eventually overtakes the line. Notice how the exponential curve appears nearly flat at first, then rockets upward after x = 3.

Two features jump out from the graph. First, both functions start at the same point (10 when x = 0), confirming they share the same initial value. Second, while the linear function climbs at a perfectly steady pace — each step adds exactly 10 — the exponential function's jumps get larger and larger. By x = 4, the exponential value has already surged past the linear value, and the gap only widens from there. This "crossing point" is a common setup for PSAT questions: they might ask you when the exponential surpasses the linear, or by how much at a specific time.

SECTION 4

Mathematical Framework

Both growth types are captured by compact equations. Knowing these forms lets you extract information quickly — slope and intercept for linear, base and initial value for exponential. On the PSAT, you may need to write, interpret, or manipulate either form.

LINEAR GROWTH MODEL

y = mx + b

y = output (value at time x) • m = slope (constant rate of change per unit of x) • b = y-intercept (starting value when x = 0)

EXPONENTIAL GROWTH MODEL

y = a · bˣ

y = output • a = initial value (when x = 0) • b = growth factor (b > 1 means growth; 0 < b < 1 means decay)

A quick way to connect the growth factor b to a percentage is to write b = 1 + r, where r is the growth rate as a decimal. For example, 20 % growth per year means r = 0.20, so b = 1.20. For decay (decreasing quantities), r is negative: a 15 % decline gives b = 1 + (−0.15) = 0.85.

GROWTH FACTOR FROM PERCENTAGE

b = 1 + r

r = growth rate as a decimal (positive for growth, negative for decay)

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PSAT Tip

SECTION 5

How to Identify Growth Type from a Table

PSAT questions often hand you a table and ask whether the relationship is linear, exponential, or neither. The fastest diagnostic is a two-step process: first check the differences between consecutive outputs, then check the ratios. If the differences are constant, it's linear. If the ratios are constant, it's exponential.

Follow this decision tree whenever a PSAT question gives you a data table. Compute differences first; if they're constant, you have a linear model. If not, compute ratios; if those are constant, you have an exponential model.

x	y (Linear)	Difference	y (Exponential)	Ratio
0	100	—	100	—
1	120	+20	120	×1.2
2	140	+20	144	×1.2
3	160	+20	172.8	×1.2
4	180	+20	207.36	×1.2

Side-by-side comparison: constant differences (linear) vs. constant ratios (exponential)

Notice that both sequences start at 100, but by x = 4 the exponential value has already pulled ahead (207.36 vs. 180). The linear column adds 20 each time; the exponential column multiplies by 1.2 each time. Even though the amounts added in the exponential column are different (20, 24, 28.8, 34.56), the ratio between consecutive values is always 1.2. That constant ratio is the hallmark of exponential behavior.

SECTION 6

Worked Example

Let's work through a PSAT-style problem from start to finish. Pay attention to how we identify the growth type, build the equation, and use it to answer the question.

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PROBLEM

Worked example

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Step 1 — Identify Growth Types

Colony A gains a constant 150 each hour → linear. Colony B grows by a constant 25 % each hour → exponential.
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Step 2 — Write the Equations

Colony A: y = 500 + 150t (linear form, m = 150, b = 500). Colony B: y = 500 × 1.25ᵗ (exponential form, a = 500, b = 1 + 0.25 = 1.25).
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Step 3 — Build a Table to Compare

t = 0: A = 500, B = 500. t = 1: A = 650, B = 625. t = 2: A = 800, B ≈ 781. t = 3: A = 950, B ≈ 977. At t = 3, Colony B (≈ 977) first exceeds Colony A (950).
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Step 4 — Verify and State Answer

Check: at t = 2, A = 800 > B ≈ 781, so B has not yet surpassed A. At t = 3, B ≈ 976.6 > A = 950.

Colony B first exceeds Colony A after 3 complete hours.

SECTION 7

Linear vs. Exponential — Side-by-Side Comparison

It helps to have a compact reference that contrasts every major feature of the two models. The table below covers the properties that show up most often on the PSAT.

Feature	Linear Growth	Exponential Growth
Equation	y = mx + b	y = a · bˣ
Graph Shape	Straight line	J-shaped curve (growth) or decaying curve
Rate of Change	Constant (slope m)	Changes — proportional to current value
Table Test	Constant differences between consecutive y-values	Constant ratios between consecutive y-values
Key Phrase Clues	"increases by 5 each…", "per year adds…"	"increases by 5 % each…", "doubles every…"
Long-Run Winner	Grows without bound, but slowly	Always overtakes linear eventually
Real-World Example	Hourly wages, constant speed travel	Compound interest, population growth

Key contrasts between linear and exponential models

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Key takeaway

LANGUAGE CUES

SECTION 8

Connecting to Advanced Topics

While the PSAT focuses on basic linear and exponential models, understanding how these ideas extend will deepen your intuition and prepare you for SAT, AP, and college-level work. The table below maps what you know now to what comes next.

PSAT Level	Advanced Extension
y = mx + b (linear)	Systems of linear equations, linear regression (line of best fit)
y = a · bˣ (exponential)	Logarithms to solve for x; continuous growth model y = a · eʳᵗ
Constant difference → linear	Constant second difference → quadratic (y = ax² + bx + c)
Growth factor b > 1	Logistic growth models with carrying capacity (AP Biology / AP Calc)

How PSAT-level concepts connect to future coursework

One particularly useful bridge is the idea of exponential decay. When the base b is between 0 and 1, the exponential function decreases over time — think of a car losing 15 % of its value each year (b = 0.85) or a medication being cleared from the bloodstream. The PSAT tests decay just as often as growth, so make sure you're comfortable with both directions.

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Looking Ahead

SECTION 9

Practice Problems

Test your understanding with these five problems, arranged from foundational to challenging. Try each one on your own before reading the answer.

Question 1: CONCEPTUAL

A town's population increases by exactly 200 people every year. Is this linear growth, exponential growth, or neither? Explain your reasoning.

Show answer

This is linear growth because the population increases by a constant amount (200) each year. The key diagnostic is that the difference between consecutive years is always the same, not the ratio. If the town grew by a constant percentage instead, it would be exponential.

Question 2: BASIC CALCULATION

A savings account starts with $2,000 and earns 5 % interest per year (compounded annually). Write the exponential equation and find the account balance after 3 years.

Show answer

The exponential model is y = 2000 × 1.05ᵗ, where t is the number of years. After 3 years: y = 2000 × 1.05³ = 2000 × 1.157625 = $2,315.25. The growth factor is b = 1 + 0.05 = 1.05. Each year, the balance is multiplied by 1.05, meaning you earn interest on both the original deposit and previously earned interest.

Question 3: INTERMEDIATE

A data table shows x-values of 0, 1, 2, 3, 4 and corresponding y-values of 6, 18, 54, 162, 486. Determine whether the relationship is linear or exponential, then write the equation.

Show answer

First, check differences: 18 − 6 = 12, 54 − 18 = 36, 162 − 54 = 108. These are not constant, so it's not linear. Next, check ratios: 18 ÷ 6 = 3, 54 ÷ 18 = 3, 162 ÷ 54 = 3, 486 ÷ 162 = 3. The ratios are all constant at 3, so this is exponential. The initial value (when x = 0) is 6, and the growth factor is 3. The equation is y = 6 × 3ˣ.

Question 4: APPLIED

A phone's battery drains at a constant rate, losing 12 % of its remaining charge each hour. If the battery starts at 100 %, what percentage of charge remains after 5 hours? Round to the nearest whole percent.

Show answer

Because the battery loses 12 % of its remaining charge each hour, this is exponential decay with b = 1 − 0.12 = 0.88. The model is y = 100 × 0.88ᵗ. After 5 hours: y = 100 × 0.88⁵ = 100 × 0.52773 ≈ 53 %. A common mistake would be saying 100 − 5(12) = 40 %, but that assumes a constant 12 percentage points lost each hour (linear), not 12 % of whatever remains.

Question 5: CRITICAL THINKING

Plan A pays $50 per day. Plan B pays 1 cent on day 1, 2 cents on day 2, 4 cents on day 3, doubling each day. On which day does Plan B's single-day payment first exceed Plan A's daily payment of $50?

A. Day 12
B. Day 13
C. Day 14
D. Day 16

Show answer

Plan A is linear: it pays a flat $50 every day. Plan B is exponential: the payment on day n equals 0.01 ×

2^{n−1}

dollars. To find when Plan B's daily payment first exceeds $50, solve: 0.01 ×

2^{n−1}

> 50, which gives

2^{n−1}

> 5,000. Testing powers of 2:

2^{12}

= 4,096 (less than 5,000) and

2^{13}

= 8,192 (greater than 5,000). Therefore n − 1 = 13, so n = 14. Verification: day 14 payment = 0.01 ×

2^{13}

= 0.01 × 8,192 = $81.92, which exceeds $50. Day 13 payment = 0.01 ×

2^{12}

= $40.96, which does not exceed $50. Therefore Plan B's single-day payment first exceeds Plan A's on day 14. This illustrates a key principle: exponential growth, even when it starts from an extremely small value, will always eventually overtake any linear function. The power of repeated doubling should never be underestimated.