SAT Math – Problem Solving & Data Analysis
Linear and Exponential Growth
Comparing constant change versus multiplicative growth patterns
Linear and exponential growth represent two fundamentally different patterns of change. On the SAT, you'll distinguish between them, model real-world situations with appropriate equations, and predict future values—skills that apply everywhere from population dynamics and compound interest to viral spread and technology adoption.
Success requires recognizing which type of growth fits a situation, understanding the structure of linear versus exponential equations, calculating growth over time, and comparing rates of change. These aren't just mathematical abstractions—they're the models that explain why epidemics spread exponentially, why savings grow through compound interest, and why some changes accelerate while others proceed steadily.
Understanding the Two Growth Types
Linear Growth
Linear growth increases (or decreases) by the same absolute amount each time period. The change is constant.
Equation: \(y = mx + b\) or \(y = a + bx\)
Example: Saving $50 per month: 50, 100, 150, 200...
Rate: Constant amount per time period
Exponential Growth
Exponential growth increases (or decreases) by the same percentage each time period. The change accelerates.
Equation: \(y = a \cdot b^x\) or \(y = a(1 + r)^x\)
Example: Growing by 10% each month: 100, 110, 121, 133.1...
Rate: Constant percentage per time period
Key Difference
The crucial distinction: linear adds, exponential multiplies.
Exponential: Same proportional increase each period
Over time: Exponential eventually grows much faster than linear
Recognition: Check if successive differences are constant (linear) or successive ratios are constant (exponential)
Essential Formulas & Concepts
Linear Growth Formula
\(y = a + bx\)
a: Initial value (starting amount)
b: Rate of change (amount added per period)
x: Number of time periods
Exponential Growth Formula
\(y = a \cdot b^x\) or \(y = a(1 + r)^x\)
a: Initial value (starting amount)
b: Growth factor (what you multiply by each period)
r: Growth rate as a decimal (b = 1 + r)
x: Number of time periods
Growth vs. Decay
Exponential Growth: \(b > 1\) or \(r > 0\)
Example: 5% growth → \(b = 1.05\)
Exponential Decay: \(0 < b < 1\) or \(-1 < r < 0\)
Example: 20% decay → \(b = 0.80\) or \(r = -0.20\)
Identifying the Pattern
For Linear: Check if differences between consecutive terms are constant
Example: 5, 8, 11, 14 → differences are 3, 3, 3 (linear)
For Exponential: Check if ratios between consecutive terms are constant
Example: 5, 10, 20, 40 → ratios are 2, 2, 2 (exponential)
Common Pitfalls & Expert Tips
❌ Confusing growth rate with growth factor
10% growth rate means multiply by 1.10, not 0.10! Growth factor = 1 + (rate as decimal). Don't forget to add 1!
❌ Using wrong formula for the pattern
If a situation describes "increases by $500 each year," that's linear. If it says "increases by 5% each year," that's exponential. Keywords matter!
❌ Forgetting the exponent in exponential formulas
Exponential growth is \(a \cdot b^x\), not \(a \cdot bx\). The exponent makes all the difference!
❌ Mishandling decay problems
For 15% decrease, use 0.85 (not 0.15). You keep 85% of the previous amount: 1 - 0.15 = 0.85.
✓ Expert Tip: Look for key phrases
"Increases by [number]" = linear. "Increases by [percent]" or "multiplies by" = exponential. Train yourself to recognize these instantly.
✓ Expert Tip: Check the pattern visually
Linear graphs are straight lines. Exponential graphs curve upward (growth) or downward (decay). If you see a curve, think exponential!
✓ Expert Tip: Time zero matters
The initial value 'a' is the amount at time x = 0. Make sure you're counting time periods correctly from the starting point.
Fully Worked SAT-Style Examples
A savings account starts with $200 and increases by $50 every month. What is the account balance after 8 months?
Solution:
Step 1: Identify the growth type
"Increases by $50 every month" = same absolute amount
This is LINEAR growth
Step 2: Set up linear equation
\(y = a + bx\)
\(a = 200\) (initial amount)
\(b = 50\) (amount added per month)
\(x = 8\) (number of months)
Step 3: Calculate
\(y = 200 + 50(8)\)
\(y = 200 + 400 = 600\)
Answer: $600
A population of bacteria starts at 500 and increases by 20% each hour. How many bacteria are there after 3 hours?
Solution:
Step 1: Identify the growth type
"Increases by 20%" = same percentage each time
This is EXPONENTIAL growth
Step 2: Determine growth factor
20% increase means multiply by 1.20
Growth factor \(b = 1 + 0.20 = 1.20\)
Step 3: Set up exponential equation
\(y = a \cdot b^x = 500 \cdot (1.20)^3\)
Step 4: Calculate
\((1.20)^3 = 1.20 \times 1.20 \times 1.20 = 1.728\)
\(y = 500 \times 1.728 = 864\)
Answer: 864 bacteria
Investment A starts at $1,000 and grows by $100 per year (linear). Investment B starts at $1,000 and grows by 8% per year (exponential). Which has a higher value after 5 years?
Solution:
Calculate Investment A (Linear):
\(y = 1000 + 100(5) = 1000 + 500 = 1500\)
Calculate Investment B (Exponential):
Growth factor = \(1 + 0.08 = 1.08\)
\(y = 1000 \cdot (1.08)^5\)
\((1.08)^5 \approx 1.469\)
\(y = 1000 \times 1.469 = 1469\)
Compare:
Investment A: $1,500
Investment B: $1,469
Investment A is higher after 5 years
Important Note:
In early years, linear can outpace exponential
But exponential eventually surpasses linear over longer time periods
Answer: Investment A ($1,500 vs $1,469)
A car purchased for $24,000 depreciates by 15% each year. What is its value after 4 years?
Solution:
Step 1: Identify decay factor
15% decrease means you keep 85% of the value
Decay factor = \(1 - 0.15 = 0.85\)
Step 2: Set up exponential decay equation
\(y = 24{,}000 \cdot (0.85)^4\)
Step 3: Calculate
\((0.85)^4 \approx 0.522\)
\(y = 24{,}000 \times 0.522 = 12{,}528\)
Common Error:
Don't use 0.15 as the factor!
15% decrease means multiply by 0.85, not 0.15
Answer: $12,528
A sequence of values is: 8, 12, 16, 20, 24. Is this linear or exponential growth? Write the equation.
Solution:
Step 1: Check differences (for linear)
12 - 8 = 4
16 - 12 = 4
20 - 16 = 4
24 - 20 = 4
Constant difference of 4 → LINEAR growth
Step 2: Write linear equation
Starting value (when \(x = 0\)): 8
Rate of change: 4 per period
\(y = 8 + 4x\)
Verification:
\(x = 0\): \(y = 8 + 4(0) = 8\) ✓
\(x = 1\): \(y = 8 + 4(1) = 12\) ✓
\(x = 2\): \(y = 8 + 4(2) = 16\) ✓
Answer: Linear growth; \(y = 8 + 4x\)
A sequence of values is: 3, 6, 12, 24, 48. Is this linear or exponential growth? Write the equation.
Solution:
Step 1: Check differences (for linear)
6 - 3 = 3
12 - 6 = 6
24 - 12 = 12
Differences are NOT constant → Not linear
Step 2: Check ratios (for exponential)
\(\frac{6}{3} = 2\)
\(\frac{12}{6} = 2\)
\(\frac{24}{12} = 2\)
\(\frac{48}{24} = 2\)
Constant ratio of 2 → EXPONENTIAL growth
Step 3: Write exponential equation
Starting value (when \(x = 0\)): 3
Growth factor: 2
\(y = 3 \cdot 2^x\)
Answer: Exponential growth; \(y = 3 \cdot 2^x\)
Maria invests $5,000 in an account that earns 6% interest compounded annually. How much will she have after 10 years?
Solution:
Step 1: Recognize compound interest as exponential
6% interest per year = exponential growth
Growth factor = \(1 + 0.06 = 1.06\)
Step 2: Set up equation
\(y = 5{,}000 \cdot (1.06)^{10}\)
Step 3: Calculate
\((1.06)^{10} \approx 1.791\)
\(y = 5{,}000 \times 1.791 = 8{,}955\)
Note on Compound Interest:
Compound interest is always exponential growth
Each year, you earn interest on the previous year's total
This creates a multiplicative (exponential) pattern
Answer: $8,955
A population follows the exponential model \(P = 10{,}000 \cdot (1.15)^t\), where t is time in years. What is the population after 5 years?
Solution:
Step 1: Understand the given equation
Initial population: 10,000
Growth factor: 1.15 (15% growth per year)
Variable: t = years
Step 2: Substitute t = 5
\(P = 10{,}000 \cdot (1.15)^5\)
Step 3: Calculate
\((1.15)^5 \approx 2.011\)
\(P = 10{,}000 \times 2.011 = 20{,}110\)
Answer: 20,110 people
Quick Reference Comparison
SAT Growth Problem Checklist
Identifying Growth Type
- Same amount added? → Linear
- Same percent increase? → Exponential
- Check differences (linear)
- Check ratios (exponential)
For Linear Problems
- Use \(y = a + bx\)
- Identify starting value (a)
- Find rate of change (b)
- Count time periods (x)
For Exponential Problems
- Use \(y = a \cdot b^x\)
- Growth: \(b = 1 + r\)
- Decay: \(b = 1 - r\)
- Don't forget the exponent!
Common Mistakes
- Confusing rate with factor
- Using wrong formula type
- Forgetting exponent notation
- Decay: use (1 - r), not r
Linear and Exponential Growth: The Mathematics of Change
Understanding the difference between linear and exponential growth is fundamental to making sense of the modern world. Linear growth describes steady, predictable change—salaries that increase by fixed amounts, distances covered at constant speed, savings that grow by regular deposits. Exponential growth describes accelerating change—populations that double, investments that compound, viral content that spreads, epidemics that surge. The SAT tests these concepts because they represent essential quantitative reasoning: recognizing which pattern fits a situation, predicting future values, and understanding that exponential processes eventually dominate linear ones. This knowledge applies everywhere—from evaluating investment options and understanding pandemic spread to recognizing unsustainable growth patterns and interpreting scientific data. Master both models not just for test success, but to become someone who can think critically about growth, change, and the mathematics underlying systems all around you. When policymakers discuss population growth, when scientists model climate change, when businesses project revenues, they're applying these same fundamental concepts of linear versus exponential change.
