Exponential Functions and Growth/Decay

Model exponential growth and decay for the Digital SAT. Understand f(x) = a·bˣ, doubling times, half-lives, and compound interest.

Exponential functions model quantities that multiply by a constant factor over equal intervals. Unlike linear functions (which add a constant), exponential functions grow or decay by a percentage. The Digital SAT tests exponential models in contexts like population growth, radioactive decay, compound interest, and depreciation.

Core Concepts

The General Exponential Function

f(x)=abxf(x) = a \cdot b^x

  • aa = initial value (when x=0x = 0, f(0)=af(0) = a)
  • bb = growth/decay factor
  • If b>1b > 1: exponential growth
  • If 0<b<10 < b < 1: exponential decay

Growth Factor vs. Growth Rate

If a quantity increases by r%r\% per period:

b=1+r100b = 1 + \frac{r}{100}

If it decreases by r%r\%:

b=1r100b = 1 - \frac{r}{100}

Example: Population grows 5% per year: b=1.05b = 1.05 Example: Car depreciates 15% per year: b=0.85b = 0.85

Compound Interest

A=P(1+rn)ntA = P\left(1 + \frac{r}{n}\right)^{nt}

  • PP = principal
  • rr = annual rate (decimal)
  • nn = compounding periods per year
  • tt = years

Continuous Growth/Decay

f(t)=aektf(t) = a \cdot e^{kt}

  • k>0k > 0: growth
  • k<0k < 0: decay

Key Properties

  • Exponential functions have a horizontal asymptote (usually y=0y = 0).
  • They never reach zero (for decay) or become negative.
  • They eventually outgrow any linear or polynomial function.

Doubling Time and Half-Life

Doubling time: how long until the quantity doubles. Set abt=2aa \cdot b^t = 2abt=2b^t = 2.

Half-life: how long until half remains. Set abt=a2a \cdot b^t = \frac{a}{2}bt=12b^t = \frac{1}{2}.

Strategy Tips

Tip 1: Identify aa and bb from Context

aa is the starting amount. bb is (1+rate)(1 + \text{rate}) or (1rate)(1 - \text{rate}).

Tip 2: Use the Table Pattern

In an exponential function, equal changes in xx produce equal ratios in yy (not equal differences — that's linear).

Tip 3: For "After How Many Years" Questions

Set up the equation and solve using logarithms or trial-and-error with the calculator.

Tip 4: Don't Confuse Linear and Exponential

Linear: constant amount added. Exponential: constant percentage multiplied.

Tip 5: Use the Calculator for Large Exponents

On the SAT, you can use the built-in calculator to evaluate expressions like 1.05201.05^{20}.

Worked Example: Example 1

Problem

A population of 2,000 bacteria triples every hour. Write the model and find the population after 4 hours.

P(t)=20003tP(t) = 2000 \cdot 3^t

P(4)=200081=162,000P(4) = 2000 \cdot 81 = 162{,}000

Solution

Worked Example: Example 2

Problem

A car worth $25,000 depreciates 12% per year. What is it worth after 5 years?

V = 25000 \cdot 0.88^5 = 25000 \cdot 0.5277 \approx \13{,}193$

Solution

Worked Example: Example 3

Problem

$5,000 is invested at 4% annual interest, compounded quarterly. Find the value after 3 years.

A = 5000\left(1 + \frac{0.04}{4}\right)^{12} = 5000(1.01)^{12} \approx \5{,}634.13$

Solution

Worked Example: SAT-Style

Problem

The equation N=500(1.08)tN = 500(1.08)^t models the number of users tt months after launch. What does 500 represent? What does 1.08 represent?

  • 500: the initial number of users at launch.
  • 1.08: the user count grows by 8% per month.
Solution

Worked Example: Example 5

Problem

A substance has a half-life of 6 hours. Starting with 100 grams, how much remains after 18 hours?

18 hours = 3 half-lives.

100(12)3=10018=12.5100 \cdot \left(\frac{1}{2}\right)^3 = 100 \cdot \frac{1}{8} = 12.5 grams.

Solution

Practice Problems

  1. Problem 1

    P(t)=1000(1.06)tP(t) = 1000 \cdot (1.06)^t. What is P(10)P(10)?

    Problem 2

    A radioactive element decays 20% per year. Starting with 500g, write the decay model.

    Problem 3

    $10,000 invested at 5% compounded annually. When does it double? (Use trial values.)

    Problem 4

    Is the function y=4(0.7)xy = 4 \cdot (0.7)^x growth or decay? What is the decay rate?

    Problem 5

    A table shows: x=0,y=3x = 0, y = 3; x=1,y=9x = 1, y = 9; x=2,y=27x = 2, y = 27. Write the equation.

    Problem 6

    The value V=20000(0.85)tV = 20000(0.85)^t models a car's value. After how many years is it worth less than $10,000?

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Common Mistakes

  • Using b=rb = r instead of b=1+rb = 1 + r. A 5% growth rate means b=1.05b = 1.05, not 0.050.05.
  • Confusing exponential with linear. If something grows by 5% (not 5 units), it's exponential.
  • Wrong exponent for compound interest. ntnt is the total number of compounding periods, not just tt.
  • Forgetting that decay factors are less than 1. A 10% decrease means multiply by 0.90.
  • Misinterpreting "tripling time" or "half-life." Set up the equation carefully.

Key Takeaways

  • f(x)=abxf(x) = a \cdot b^x where aa = initial value, bb = growth/decay factor.

  • b>1b > 1: growth. 0<b<10 < b < 1: decay.

  • Growth rate rr: b=1+rb = 1 + r. Decay rate: b=1rb = 1 - r.

  • Compound interest: A=P(1+r/n)ntA = P(1 + r/n)^{nt}.

  • Exponential functions always have a horizontal asymptote.

  • Use context to interpret aa (starting value) and bb (rate factor).

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