9.1 Exponential Functions, Growth, and Decay

9.1 Exponential Functions, Growth, and Decay

College Board’s Next-Generation ACCUPLACER Advanced Algebra and Functions (AAF) test weights exponential and logarithmic equations at 5–15% of the 20-item computer-adaptive test — typically 1–3 items. Table 11 names four skills in this content area: creating, solving, graphing, and interpreting exponential and logarithmic equations and functions. This section is the modeling core: write y = a b^t, tell a growth factor from a growth rate, convert the time unit so the exponent counts the right number of periods, and contrast the model with a linear y = mx + b.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

The model y = a b^t

An exponential function multiplies the current amount by a constant growth factor b once per time period. The template is:

y = a b^t

• a is the initial value — the amount when t = 0, because b^0 = 1 for b > 0.
• b is the growth factor per one unit of t. You need b > 0 and b ≠ 1 (if b = 1 the amount never changes; if b ≤ 0 the real-valued model is not the AAF template).
• t is the number of periods, not automatically “years” unless you defined it that way.

If b > 1, the quantity grows. If 0 < b < 1, it decays. The same template covers both; only the factor changes. AAF will also write f(x) = a b^x or P(t) = a(1 + r)^t. Those are the same family. The letter is not the model.

Growth factor versus growth rate

The rate r is the percent change per period as a decimal. The factor is what you actually multiply by:

b = 1 + r for growth, and b = 1 − r for decay when r is the fraction lost per period.

Doubling means b = 2, not b = 1.02 and not b = 100. A 2% increase would be b = 1.02. Students who treat “double” as “add 2” or as “2%” write a linear or a tiny-growth model and miss the item. Half-life means b = 1/2 per half-life, not b = −1/2 and not subtracting half the original from a fixed starting value every time.

Converting the time unit

The exponent must count how many growth periods have elapsed, not the raw clock reading. If the quantity doubles every 4 hours and the input x is measured in days, there are 24 hours in a day, so there are 24/4 = 6 doubling periods per day. After x days the exponent is 6x:

y = a · 2^{6x}

Check at one day: x = 1 should include 6 doublings, so the amount is a · 2^6 = 64a. That matches six successive doublings in 24 hours. The wrong models y = a · 2^{4x} (treats 4 as a count of periods per day), y = a · 2^{x/4} (that form is for “doubles every 4 days when x is in days”), and y = a · 2^{24x} (counts hours, not 4-hour blocks).

If instead a wildlife refuge herd doubles every 4 years and t is in years, the exponent is t/4:

E(t) = 240 · 2^{t/4}

Same algebra, different unit. Always compute (clock time) / (length of one period) or (periods per clock unit) × (clock units).

Worked example: livestream views doubling every 4 hours

A livestream opens with 50 viewers. The viewer count doubles every 4 hours. Let x be time in days after the start. Write a model for the viewer count V.

1. Initial value a = 50.
2. Growth factor per 4-hour period is b = 2.
3. Periods per day: 24/4 = 6, so after x days there are 6x doublings.
4. Model: V(x) = 50 · 2^{6x}.

Evaluate at 8 hours, which is x = 8/24 = 1/3 day: V(1/3) = 50 · 2^{6 · 1/3} = 50 · 2^2 = 200. Directly: 8 hours is two 4-hour periods, so two doublings, 50 → 100 → 200. The converted exponent and the period-count agree. If the stem instead measured t in hours, the model would be V(t) = 50 · 2^{t/4}. Same viewers, different input unit. AAF loves that conversion.

Creating an exponential from a percent rate

Worked: coastal town. A coastal town has 18,000 residents at t = 0. The population grows by 4% per year. Create a function for the population after t years.

Percent growth 4% ⇒ r = 0.04 ⇒ factor b = 1.04. Initial value 18,000.

P(t) = 18000(1.04)^t

That is creating an exponential function from a context — a Table 11 skill. Do not write 18000 + 0.04t (linear add of 0.04 people), 18000(0.04)^t (that decays toward 0 because 0.04 < 1), or 18000(4)^t (that uses 4 as a factor, which would be 300% growth).

Evaluate at t = 0: P(0) = 18000(1.04)^0 = 18000. Evaluate at t = 1: P(1) = 18000(1.04) = 18,720. Skills Insight 200–236 is exactly this: evaluate a simple exponential (or read what the value means) at a small whole number. After 3 years, P(3) = 18000(1.04)^3 = 18000(1.124864) ≈ 20,248. You do not need a calculator for the structure of the answer; many AAF options leave (1.04)^3 unexpanded.

Worked: certificate of deposit. A CD starts at $2,500 and earns 6% interest compounded annually. After t years:

A(t) = 2500(1.06)^t

After 2 years: A(2) = 2500(1.06)^2 = 2500(1.1236) = 2,809. The 1.1236 is 1.06 × 1.06, two factors, not 1.12. Compounding is multiplication of factors, not adding 6% of the original twice (that would be simple interest 2500 + 2500(0.06)(2) = 2,800, a linear model in t). Interpreting “compounded annually” means one multiplication by 1.06 per year.

Linear versus exponential

A linear model y = mx + b adds the same amount m each period. An exponential model y = a b^t multiplies by the same factor b each period. The 237–249 band connects tables to nonlinear relationships, so it expects you to look at a table or a sentence and name which family it is, then write the matching formula.

Start both families at 100:

Through year 1 the tables match, which is why a two-point glance can fool you. By year 2 they split: linear is still adding 50, exponential has added 75. By year 4 the exponential is more than 200 above the line. If a stem says “increases by 50 each year,” write linear. If it says “increases by 50% each year,” write exponential with b = 1.5. The words by 50 versus by 50% are the whole item.

A first-difference test: subtract consecutive y-values. Constant differences ⇒ linear. A ratio test: divide consecutive y-values. Constant ratios ⇒ exponential. For the exponential column, 150/100 = 1.5, 225/150 = 1.5, 337.5/225 = 1.5. For the linear column the ratios shrink toward 1 (150/100 = 1.5, then 200/150 ≈ 1.33, then 250/200 = 1.25).

Linear modeling in context is developed in Modeling with Linear Functions. On AAF exponential items, the trap is using that linear skill when the story is multiplicative. Function notation f(t) = 100(1.5)^t is the same model; evaluating f(3) is the 200–236 skill, and naming the family from a table is the 237–249 table-connection skill.

Half-life: b = 1/2

Half-life is the time in which a quantity is multiplied by 1/2. If a medical tracer starts at 64 mg and has half-life 8 hours, then after t hours the number of half-lives is t/8, so:

M(t) = 64 · (1/2)^{t/8}

Evaluate M(24) = 64 · (1/2)^3 = 64/8 = 8. Skills Insight 200–236 may ask only for that evaluation, not for a half-life formula derivation.

Traps: 64 · 2^{t/8} is growth, the inverse of decay. 64 − 8t is linear subtraction and hits 0 at 8 hours, which is not half-life. 64 · (1/2)^{8t} treats 8 as if it belonged in the numerator of the exponent, so at t = 1 hour you would already have applied eight half-lives. Write b = 1/2 per half-life, then put (elapsed time)/(half-life) in the exponent.

If 15% of a pollutant is removed each month, that is not half-life. It is decay with b = 0.85: Q(t) = Q_0 (0.85)^t. Half-life is specifically factor 1/2. After two half-lives you have one-fourth of the original, not zero.

Interpreting a and b from a table

A table that starts at 80 and multiplies by 1.25 each step is y = 80(1.25)^t if t counts those steps from 0. If the table starts at t = 2 already, back up: divide by 1.25 twice to recover a. Creating the function from data is the same Table 11 “creating” skill as the town and the CD.

If two points are (0, 12) and (3, 96) and the model is exponential, then a = 12 and 12 b^3 = 96, so b^3 = 8, b = 2, hence y = 12 · 2^t. Check: at t = 3, 12 · 8 = 96. A linear fit through the same two points would be a different function and would miss any third exponential point.

Handheld calculators are not allowed on AAF except with an approved accommodation (College Board calculator policy). Some items show an on-screen calculator icon; many exponential-evaluation items do not. For 2^5, 3^4, and (1/2)^3, compute by hand. Leave answers such as (log_3 11)/4 for Solving Exponential and Logarithmic Equations. Graphs of y = a^x are next, in Graphs of Exponential Functions.