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5.2 Exponential Functions: Growth & Decay Models

Key Takeaways

  • The standard exponential model is f(t) = a · b^t = a(1 ± r)^t, where a = f(0) is the initial value, b is the base growth/decay factor, and r = |b - 1| is the percent rate of change per unit time.

  • Differentiate growth from decay: if b > 1, the model represents exponential growth with rate r = b - 1; if 0 < b < 1, it represents exponential decay with rate r = 1 - b.

  • Perform time-scale conversions: rewrite annual models to monthly models using f(t) = a · (b^{1/12})^{12t}, or models that scale every k time units using b^{t/k}.

  • Interpret constants in contextual SAT questions without calculating by directly mapping initial quantities, percentage growth rates, and elapsed unit intervals.

  • Distinguish linear models (constant absolute change per unit time, f(t) = mt + d) from exponential models (constant percentage or multiplicative factor change per unit time).

Last updated: August 2026

Exponential Functions: Growth & Decay Models

Exponential functions model real-world phenomena where a quantity changes by a constant percentage or constant multiplicative factor over equal intervals of time. On the Digital SAT, exponential modeling questions test your ability to construct equations, interpret constants in real-world contexts, and convert between different time scales.


1. Anatomy of the Exponential Model

The standard formula for an exponential function is written in one of two equivalent forms:

f(t)=a⋅btorf(t)=a(1±r)tf(t) = a \cdot b^t \quad \text{or} \quad f(t) = a(1 \pm r)^t
+-----------------------------------------------------------------------------+
|                     ANATOMY OF AN EXPONENTIAL MODEL                         |
|                                                                             |
|                             f(t) = a · b^t                                  |
|                                    |   |                                    |
|             +----------------------+   +-----------------------+            |
|             v                                                  v            |
|     [ INITIAL VALUE (a) ]                             [ GROWTH FACTOR (b) ] |
|     - Value when t = 0                                - b = 1 + r (Growth)  |
|     - y-intercept on graph: (0, a)                    - b = 1 - r (Decay)   |
|     - Starting population / principal                 - b > 1: Increases    |
|                                                       - 0 < b < 1: Decreases|
+-----------------------------------------------------------------------------+

Parameter Breakdown:

  1. Initial Value (aa): The output when t=0t = 0, represented by the point (0,a)(0, a) on the Cartesian plane. In context, aa represents the starting population, initial investment, original purchase price, or baseline quantity.
  2. Base / Growth Factor (bb): The multiplier applied for each single unit increase in tt.
    • If b>1b > 1, the function models Exponential Growth.
    • If 0<b<10 < b < 1, the function models Exponential Decay.
    • bb can never be negative or zero in real-world exponential models.
  3. Percent Rate of Change (rr): The fractional or percentage increase/decrease per unit time:
    • Growth: b=1+r  ⟹  r=b−1b = 1 + r \implies r = b - 1
    • Decay: b=1−r  ⟹  r=1−bb = 1 - r \implies r = 1 - b
  4. Independent Variable (tt): The number of elapsed time periods (e.g., years, months, days, hours).

2. Growth Factor vs. Decay Factor Conversion

The table below demonstrates how percentage changes translate into exponential base values (bb):

Context DescriptionPercent Change (rr)Base FormulaFactor (bb)Model Equation (a=500a = 500)
Grows by 8% per year+8%=+0.08+8\% = +0.08b=1+0.08b = 1 + 0.081.081.08f(t)=500(1.08)tf(t) = 500(1.08)^t
Decreases by 15% per year−15%=−0.15-15\% = -0.15b=1−0.15b = 1 - 0.150.850.85f(t)=500(0.85)tf(t) = 500(0.85)^t
Grows by 0.6% per month+0.6%=+0.006+0.6\% = +0.006b=1+0.006b = 1 + 0.0061.0061.006f(t)=500(1.006)tf(t) = 500(1.006)^t
Depreciates by 4.2% per year−4.2%=−0.042-4.2\% = -0.042b=1−0.042b = 1 - 0.0420.9580.958f(t)=500(0.958)tf(t) = 500(0.958)^t
Doubles every year+100%=+1.00+100\% = +1.00b=1+1.00b = 1 + 1.002.002.00f(t)=500(2)tf(t) = 500(2)^t
Triples every year+200%=+2.00+200\% = +2.00b=1+2.00b = 1 + 2.003.003.00f(t)=500(3)tf(t) = 500(3)^t
Halves every year−50%=−0.50-50\% = -0.50b=1−0.50b = 1 - 0.500.500.50f(t)=500(0.5)tf(t) = 500(0.5)^t

Important

Common Percent-to-Decimal Traps:

  • A 5%5\% increase is b=1+0.05=1.05b = 1 + 0.05 = 1.05, NOT 1.51.5 (which is a 50%50\% increase).
  • A 0.4%0.4\% increase is b=1+0.004=1.004b = 1 + 0.004 = 1.004, NOT 1.041.04 (which is a 4%4\% increase).
  • A 30%30\% decay means 70%70\% remains (b=0.70b = 0.70), NOT b=0.30b = 0.30 (which would mean a 70%70\% decay).

3. Time-Scale Transformations & Exponent Rewriting

The Digital SAT frequently tests equivalent exponential equations where the time unit is modified. These questions rely on the Power of a Power property: bt=(b1/k)kt=(bk)t/kb^t = (b^{1/k})^{kt} = (b^k)^{t/k}.

+-----------------------------------------------------------------------------+
|                    TIME-SCALE TRANSFORMATION METHODS                        |
|                                                                             |
|   ANNUAL TO MONTHLY CONVERSION:                                             |
|   f(t) = a · (1 + r)^t  ====>  f(t) = a · [ (1 + r)^(1/12) ]^(12t)          |
|                                          -----------------                  |
|                                            Monthly Base                     |
|                                                                             |
|   PERIODIC CHANGE (EVERY k YEARS):                                          |
|   f(t) = a · b^(t / k)                                                      |
|   - When t = k, exponent = 1 (one multiplication by b)                      |
|   - When t = 2k, exponent = 2 (two multiplications by b)                    |
+-----------------------------------------------------------------------------+

Case 1: Converting from Annual Rate to Monthly Rate

Suppose an investment grows by 12%12\% per year: f(t)=1,000(1.12)tf(t) = 1,000(1.12)^t, where tt is years. To find the monthly growth factor:

  1. There are 1212 months in a year, so elapsed time in months is m=12tm = 12t, which means t=m12t = \frac{m}{12}.
  2. Substitute t=m12t = \frac{m}{12}: f(m)=1,000(1.12)m/12=1,000((1.12)1/12)mf(m) = 1,000(1.12)^{m/12} = 1,000\left((1.12)^{1/12}\right)^m.
  3. Calculate the monthly base: (1.12)1/12≈1.009488(1.12)^{1/12} \approx 1.009488.
  4. This reveals a monthly growth rate of approximately 0.95%0.95\% per month.

Case 2: Periodic Change Every kk Units

If a population doubles every 55 years:

P(t)=P0⋅2t5P(t) = P_0 \cdot 2^{\frac{t}{5}}
  • When t=5t = 5, exponent is 5/5=1  ⟹  P(5)=P0(2)1=2P05/5 = 1 \implies P(5) = P_0(2)^1 = 2P_0.
  • When t=10t = 10, exponent is 10/5=2  ⟹  P(10)=P0(2)2=4P010/5 = 2 \implies P(10) = P_0(2)^2 = 4P_0.
  • When t=15t = 15, exponent is 15/5=3  ⟹  P(15)=P0(2)3=8P015/5 = 3 \implies P(15) = P_0(2)^3 = 8P_0.

4. Linear vs. Exponential: The Critical SAT Distinction

SAT questions frequently present tables or verbal scenarios and ask whether the data is best modeled by a linear or exponential function:

CharacteristicLinear Model (f(t)=mt+bf(t) = mt + b)Exponential Model (f(t)=a⋅btf(t) = a \cdot b^t)
Rate of ChangeConstant absolute amount added/subtracted per unit timeConstant percentage / ratio multiplied per unit time
First DifferencesEqual differences: yk+1−yk=my_{k+1} - y_k = mIncreasing or decreasing differences
Ratios of Consecutive TermsChanging ratiosConstant ratio: yk+1yk=b\frac{y_{k+1}}{y_k} = b
Example Keywords"increases by $50 per year", "loses 3 gallons per hour""increases by 5% per year", "doubles every decade"
Graph ShapeStraight line with constant slopeCurved line with horizontal asymptote

5. Step-by-Step Worked SAT Exam Examples

Worked Example 1: Interpreting Exponential Parameters

Problem: The value of a high-performance computer server, in dollars, is modeled by V(t)=4,200(0.82)t/2V(t) = 4,200(0.82)^{t/2}, where tt is the number of years after purchase. Which statement best interprets the number 0.820.82 in this context?

Step-by-Step Solution:

  1. Identify the base: b=0.82b = 0.82. Because 0.82<10.82 < 1, this represents exponential decay.
  2. Calculate the decay percentage: r=1−0.82=0.18=18%r = 1 - 0.82 = 0.18 = 18\%.
  3. Inspect the exponent: The exponent is t2\frac{t}{2}, which indicates that one full decay cycle occurs every time tt increases by 22.
  4. Conclusion: The value of the server decreases by 18%18\% every 22 years (or retains 82%82\% of its value every 22 years).

Worked Example 2: Constructing an Equivalent Monthly Model

Problem: A tech company's active user base U(t)U(t), in thousands, is modeled by U(t)=150(1.24)tU(t) = 150(1.24)^t, where tt is the number of years since 2022. Which equivalent function reveals the approximate monthly growth rate as a constant base?

Step-by-Step Solution:

  1. There are 1212 months in a year, so if mm is the number of elapsed months then m=12tm = 12t and t=m12t = \frac{m}{12}.
  2. Rewrite the expression using exponent properties: U(t)=150⋅(1.24)t=150⋅((1.24)1/12)12tU(t) = 150 \cdot (1.24)^t = 150 \cdot \left((1.24)^{1/12}\right)^{12t}
  3. Evaluate (1.24)1/12(1.24)^{1/12} using a calculator: (1.24)1/12≈1.018087≈1.0181(1.24)^{1/12} \approx 1.018087 \approx 1.0181.
  4. Substitute back into the equation: U(t)≈150(1.0181)12tU(t) \approx 150(1.0181)^{12t} This indicates an approximate monthly growth rate of 1.81%1.81\% per month.

6. Desmos Testing Strategies & Rapid Visual Validation

Tip

Testing Equivalent Exponential Forms in Desmos: When an SAT problem asks which rewritten exponential equation is equivalent to an original annual model (e.g., f(x)=150(1.24)xf(x) = 150(1.24)^x vs. g(x)=150(1.0181)12xg(x) = 150(1.0181)^{12x}):

  1. Graph Line 1: y = 150 * (1.24)^x
  2. Graph Line 2: y = 150 * (1.0181)^(12x)
  3. Click the gear icon to convert both to tables. Inspect values at x=0,1,2,5x = 0, 1, 2, 5.
  4. If the outputs match at all test values, the models are equivalent.

Note

Exponential Regression (y1∼a⋅bx1y_1 \sim a \cdot b^{x_1}): If a problem provides a table of values (x1,y1)(x_1, y_1) and asks for the exponential equation that models them:

  1. Click the + button in Desmos and insert a table. Enter the (x1,y1)(x_1, y_1) coordinate pairs.
  2. On a new line, type y1 ~ a * b^(x1).
  3. Desmos will compute the exact parameters aa and bb instantly.
Test Your Knowledge

The population of a city is modeled by the function P(t)=14,500(1.035)tP(t) = 14,500(1.035)^t, where tt represents the number of years after 2010. By approximately what percentage does the population grow every decade (10 years)?

A

3.5%3.5\%

B

35.0%35.0\%

C

38.2%38.2\%

D

41.1%41.1\%

Test Your Knowledge

A sample of a radioactive element decays according to the equation M(t)=800(0.88)t/3M(t) = 800(0.88)^{t/3}, where M(t)M(t) is the remaining mass in grams after tt hours. Which statement correctly interprets the value 0.880.88 in this model?

A

The mass of the sample decreases by 12%12\% every 33 hours.

B

The mass of the sample decreases by 88%88\% every 33 hours.

C

The mass of the sample decreases by 12%12\% every 11 hour.

D

The mass of the sample decreases by 88%88\% every 11 hour.

Test Your Knowledge

The estimated value of a vintage guitar is modeled by V(t)=2,000(1.06)tV(t) = 2,000(1.06)^t, where tt is the number of years since appraisal. Which of the following equivalent equations correctly displays the approximate monthly growth rate as a base constant?

A

V(t)=2,000(0.088)12tV(t) = 2,000(0.088)^{12t}

B

V(t)=2,000(1.005)t/12V(t) = 2,000(1.005)^{t/12}

C

V(t)≈2,000(1.00487)12tV(t) \approx 2,000(1.00487)^{12t}

D

V(t)≈2,000(1.72)12tV(t) \approx 2,000(1.72)^{12t}

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