Exponential Functions

Master exponential growth and decay, interpret exponential expressions, solve exponential equations, and model real-world phenomena with exponential functions.

๐ŸŽฏโญ INTERACTIVE LESSON

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Exponential Functions and Equations on the SAT

What Is an Exponential Function?

An exponential function has the form: f(x)=aโ‹…bxf(x) = a \cdot b^x

Where:

  • aa = initial value (when x=0x = 0)
  • bb = base (growth/decay factor)
  • xx = typically time or number of periods

Growth vs. Decay

| Condition | Type | Example | |---|---|---| | b>1b > 1 | Exponential Growth | Population doubling | | 0<b<10 < b < 1 | Exponential Decay | Radioactive decay | | b=1b = 1 | No change (constant) | โ€” |

Growth Rate and Decay Rate

If value increases by r%r\% per period: f(x)=a(1+r)xf(x) = a(1 + r)^x

If value decreases by r%r\% per period: f(x)=a(1โˆ’r)xf(x) = a(1 - r)^x

Example: A car worth $25,000 depreciates by 15% per year: V(t)=25000(1โˆ’0.15)t=25000(0.85)tV(t) = 25000(1 - 0.15)^t = 25000(0.85)^t


Identifying Exponential Functions

| Feature | Linear | Exponential | |---|---|---| | Pattern | Add a constant | Multiply by a constant | | Equation | y=mx+by = mx + b | y=aโ‹…bxy = a \cdot b^x | | Rate of change | Constant | Changes (accelerating) | | Graph | Straight line | Curved |

From a Table

| xx | yy (Linear) | yy (Exponential) | |---|---|---| | 0 | 5 | 5 | | 1 | 8 | 10 | | 2 | 11 | 20 | | 3 | 14 | 40 |

Linear: +3+3 each time. Exponential: ร—2\times 2 each time.


Compound Interest

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

| Variable | Meaning | |---|---| | AA | Final amount | | PP | Principal (initial) | | rr | Annual interest rate (decimal) | | nn | Times compounded per year | | tt | Time in years |

Special case โ€” continuous compounding: A=PertA = Pe^{rt}


Half-Life and Doubling Time

Half-life: A=A0(12)t/hA = A_0 \left(\frac{1}{2}\right)^{t/h} where hh is the half-life

Doubling time: A=A0โ‹…2t/dA = A_0 \cdot 2^{t/d} where dd is the doubling time


SAT Question Types

Type 1: Interpret the Function

"In f(t)=500(1.03)tf(t) = 500(1.03)^t, what does the 500 represent? What does 1.03 represent?"

  • 500 = initial value
  • 1.03 = 3% growth per period (b=1+r=1+0.03b = 1 + r = 1 + 0.03)

Type 2: Growth/Decay Identification

"Does y=200(0.85)xy = 200(0.85)^x represent growth or decay, and by what percent?"

  • Decay (because 0.85<10.85 < 1)
  • Rate = 1โˆ’0.85=0.15=15%1 - 0.85 = 0.15 = 15\% decay per period

Type 3: Evaluate at a Specific Time

"If P(t)=1000(1.05)tP(t) = 1000(1.05)^t, what is P(10)P(10)?" P(10)=1000(1.05)10โ‰ˆ1628.89P(10) = 1000(1.05)^{10} \approx 1628.89

Type 4: Compound Interest

"$5,000 at 4% annual interest compounded quarterly for 3 years" A=5000(1+0.04/4)4โ‹…3=5000(1.01)12A = 5000(1 + 0.04/4)^{4 \cdot 3} = 5000(1.01)^{12}


Common SAT Mistakes

  1. Confusing growth rate with growth factor: 5% growth has factor 1.051.05, not 0.050.05
  2. Forgetting the initial value: f(0)=af(0) = a, the initial value is the coefficient
  3. Using the wrong formula for compounding: Check what nn is (monthly = 12, quarterly = 4, etc.)
  4. Confusing linear and exponential โ€” check if the table adds or multiplies
  5. Not recognizing half-life pattern: Multiply by 12\frac{1}{2} each half-life period

๐Ÿ“š Practice Problems

1Problem 1easy

โ“ Question:

A bacteria colony starts with 100 bacteria and doubles every hour. Write a function for the population PP after tt hours, and find the population after 5 hours.

๐Ÿ’ก Show Solution

Step 1: Identify the components:

  • Initial value: a=100a = 100
  • Growth factor: b=2b = 2 (doubling)

Function: P(t)=100โ‹…2tP(t) = 100 \cdot 2^t

Step 2: Find P(5)P(5): P(5)=100โ‹…25=100โ‹…32=3,200P(5) = 100 \cdot 2^5 = 100 \cdot 32 = 3{,}200

Answer: P(t)=100โ‹…2tP(t) = 100 \cdot 2^t; after 5 hours there are 3,200 bacteria.

2Problem 2medium

โ“ Question:

The value of a car is modeled by V(t)=30000(0.82)tV(t) = 30000(0.82)^t, where tt is in years. What is the annual depreciation rate, and what will the car be worth after 3 years?

๐Ÿ’ก Show Solution

Step 1: Identify the decay rate. The base is 0.820.82, so: r=1โˆ’0.82=0.18=18%r = 1 - 0.82 = 0.18 = 18\%

The car depreciates by 18% per year.

Step 2: Find V(3)V(3): V(3)=30000(0.82)3=30000(0.551368)โ‰ˆ$16,541V(3) = 30000(0.82)^3 = 30000(0.551368) \approx \$16{,}541

Answer: 18% annual depreciation; worth approximately $16,541 after 3 years.

3Problem 3medium

โ“ Question:

Which function represents a quantity that increases by 7% each month?

A) f(x)=100(7)xf(x) = 100(7)^x B) f(x)=100(1.7)xf(x) = 100(1.7)^x C) f(x)=100(0.07)xf(x) = 100(0.07)^x D) f(x)=100(1.07)xf(x) = 100(1.07)^x

๐Ÿ’ก Show Solution

Key: "Increases by 7% each month" means growth rate r=0.07r = 0.07.

The growth factor is b=1+r=1+0.07=1.07b = 1 + r = 1 + 0.07 = 1.07.

A) b=7b = 7 โ†’ this is 600% growth, not 7% โœ— B) b=1.7b = 1.7 โ†’ this is 70% growth, not 7% โœ— C) b=0.07b = 0.07 โ†’ this is decay (and extreme decay at that) โœ— D) b=1.07b = 1.07 โ†’ this is 7% growth โœ“

Answer: D) f(x)=100(1.07)xf(x) = 100(1.07)^x

Common mistake: Using 0.070.07 or 77 as the base instead of 1.071.07.

4Problem 4hard

โ“ Question:

$2,000 is invested at 6% annual interest, compounded monthly. What is the value after 5 years?

๐Ÿ’ก Show Solution

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

Values:

  • P=2000P = 2000 (principal)
  • r=0.06r = 0.06 (6% as decimal)
  • n=12n = 12 (monthly compounding)
  • t=5t = 5 (years)

Step 1: Substitute: A=2000(1+0.0612)12ร—5A = 2000\left(1 + \frac{0.06}{12}\right)^{12 \times 5} A=2000(1+0.005)60A = 2000(1 + 0.005)^{60} A=2000(1.005)60A = 2000(1.005)^{60}

Step 2: Calculate: (1.005)60โ‰ˆ1.34885(1.005)^{60} \approx 1.34885 Aโ‰ˆ2000ร—1.34885โ‰ˆ$2,697.70A \approx 2000 \times 1.34885 \approx \$2{,}697.70

Answer: Approximately $2,697.70

SAT Tip: Make sure to identify nn correctly: monthly = 12, quarterly = 4, semiannually = 2, annually = 1.

5Problem 5expert

โ“ Question:

A radioactive substance has a half-life of 8 days. If you start with 500 grams, how much remains after 20 days?

๐Ÿ’ก Show Solution

Half-life formula: A=A0(12)t/hA = A_0 \left(\frac{1}{2}\right)^{t/h}

Values:

  • A0=500A_0 = 500 grams
  • h=8h = 8 days (half-life)
  • t=20t = 20 days

Substitute: A=500(12)20/8=500(12)2.5A = 500\left(\frac{1}{2}\right)^{20/8} = 500\left(\frac{1}{2}\right)^{2.5}

Calculate: (12)2.5=(12)2โ‹…(12)0.5=14โ‹…12=142โ‰ˆ0.17678\left(\frac{1}{2}\right)^{2.5} = \left(\frac{1}{2}\right)^2 \cdot \left(\frac{1}{2}\right)^{0.5} = \frac{1}{4} \cdot \frac{1}{\sqrt{2}} = \frac{1}{4\sqrt{2}} \approx 0.17678

Aโ‰ˆ500ร—0.17678โ‰ˆ88.4ย gramsA \approx 500 \times 0.17678 \approx 88.4 \text{ grams}

Answer: Approximately 88.4 grams

Quick check: After 8 days: 250g. After 16 days: 125g. After 24 days: 62.5g. So after 20 days (between 16 and 24), the answer should be between 62.5 and 125. โœ“