Big Idea
Exponential functions model situations where a quantity changes by the same multiplicative factor over equal time intervals.
The basic formula is simple. The tricky part is usually:
- • figuring out which growth factor matches which time unit
- • changing a growth factor from one unit to another
- • deciding whether a statement about a model is actually true
- • reading the meaning of the numbers in the equation correctly
General Exponential Model
$$f(t) = a \cdot b^t$$
$a$ = initial amount
$b$ = growth factor per unit of time
$t$ = number of time units
Growth
$b > 1$
Decay
$0 < b < 1$
1. Key Definitions
Initial amount
The value when $t = 0$. In $f(t) = a \cdot b^t$, the initial amount is $a$ because $f(0) = a \cdot b^0 = a \cdot 1 = a$.
Growth factor
How the quantity changes in one time unit.
- • $b = 1.12$ → multiplied by $1.12$ each time unit
- • $b = 0.8$ → multiplied by $0.8$ each time unit
- • $b = 2$ → doubles each time unit
Growth rate
The percent increase or decrease.
- • If $b = 1 + r$, then $r$ is the growth rate → $1.08$ means $8\%$ growth
- • If $b = 1 - r$, then $r$ is the decay rate → $0.93$ means $7\%$ decay
Time unit
The exponent only makes sense when you know the time unit. A factor of $1.05$ per day is very different from $1.05$ per week.
The time unit is where many mistakes happen. Always check what $t$ counts — days, hours, weeks, months, or something else.
2. Core Rules and Properties
Rule 1: Equal time intervals → equal multiplication
If a quantity grows exponentially, every equal time step multiplies by the same factor.
If $f(t) = 500 \cdot 1.3^t$, each 1-unit increase in $t$ multiplies the output by $1.3$.
Rule 2: Shorter interval → use a root
If you know the factor for a longer interval and want the factor for a shorter interval:
$$\text{factor per small interval} = k^{1/n}$$
where $k$ is the factor over $n$ smaller intervals
Example: Something triples every 4 hours → hourly factor is $3^{1/4}$, because multiplying that factor 4 times must give $3$.
Rule 3: Longer interval → use a power
If you know the factor for a shorter interval and want the factor for a longer interval:
Factor per day is $1.06$ → factor per 5 days is $(1.06)^5$
Rule 4: Exponents track repeated multiplication
A model like $P(t) = 200 \cdot (1.04)^t$ does not mean "add 4 each time." It means multiply by 1.04 each time.
That is why exponential growth speeds up over time.
3. Interpretation in Words
When you read an exponential model, translate each piece into a sentence.
For $N(t) = 750 \cdot (1.15)^t$:
- • the starting value is $750$
- • the quantity grows by a factor of $1.15$ each time unit
- • that means a $15\%$ increase each time unit
That verbal translation is one of the best ways to avoid errors on any exponential problem.
4. Example: Cell Culture
A culture starts with $4{,}800$ cells and triples every 6 hours. Write a function for the number of cells $h$ hours after the start.
Solution
Step 1: Initial amount
$$a = 4800$$
Step 2: Growth factor per hour
The population triples in 6 hours, so the factor per hour is:
$$3^{1/6}$$
Step 3: Write the model
$$f(h) = 4800 \cdot \left(3^{1/6}\right)^h$$
Equivalently: $f(h) = 4800 \cdot 3^{h/6}$
Check
After 6 hours: $f(6) = 4800 \cdot \left(3^{1/6}\right)^6 = 4800 \cdot 3 = 14{,}400$ ✓
5. Example: Interpreting a Model
A population is modeled by $P(d) = 12 \cdot (1.09)^d$, where $P(d)$ is in thousands and $d$ is in days. Interpret the model.
Solution
Initial amount
$P(0) = 12$
Since the output is in thousands: $12{,}000$
Daily growth
Factor: $1.09$
Rate: $9\%$ per day
In words: "The colony starts at $12{,}000$ and increases by $9\%$ each day."
6. Common Patterns in Tricky Problems
Many exponential questions are really testing whether you can match the factor to the time interval.
Look for phrases like:
Those clues tell you whether to use the given factor directly, a root, a power, or careful unit interpretation.
7. Example: Changing Time Units
A chemical sample decays with a factor of $0.49$ every 2 hours. What is the factor every 1 hour?
Solution
Let the hourly factor be $r$. Two 1-hour intervals make 2 hours:
$$r^2 = 0.49$$
$$r = \sqrt{0.49} = 0.7$$
The factor every hour is $0.7$.
Quick check
$0.7 \cdot 0.7 = 0.49$ ✓ — matches the given information.
8. Mental Checks and Estimation
Quick estimates tell you whether an answer makes sense before you do the full calculation.
Check 1: Growth or decay?
- • Increasing quantity → factor must be greater than $1$
- • Decreasing quantity → factor must be between $0$ and $1$
So if a population doubles, a factor like $\frac{1}{\sqrt[5]{2}}$ cannot be right — it is less than $1$.
Check 2: Shorter intervals → factor moves closer to $1$
Doubling over 8 days → daily factor is $2^{1/8}$, which is greater than $1$ but much closer to $1$ than $2$ is. Each single day only causes a smaller increase.
Check 3: Longer intervals → factor moves farther from $1$
Factor of $1.03$ per hour → over 10 hours the factor is $(1.03)^{10}$, which is bigger than $1.03$.
9. Example: Depreciation
A machine part loses value exponentially. Its value is multiplied by $0.81$ every 3 years. Write a model for its value $V(t)$ after $t$ years if the initial value is $$9{,}500$.
Solution
Step 1: Initial amount
$$a = 9500$$
Step 2: Yearly factor
The factor for 3 years is $0.81$, so the factor for 1 year is:
$$0.81^{1/3}$$
Step 3: Write the model
$$V(t) = 9500 \cdot \left(0.81^{1/3}\right)^t$$
Equivalently: $V(t) = 9500 \cdot (0.81)^{t/3}$
Check
After 3 years: $V(3) = 9500 \cdot (0.81)^{3/3} = 9500 \cdot 0.81$ ✓
10. Reading True Statements from a Model
Many questions ask which statements are true about a model. The safest way is to test each statement directly.
Let $M(w) = 18 \cdot (1.05)^w$ represent the number of visitors to a site, in thousands, after $w$ weeks. Decide whether each statement is true.
The site started with $18{,}000$ visitors
$M(0) = 18$, and the model is in thousands.
The growth factor per week is $1.05$
That is the base of the exponential.
The growth rate per week is $5\%$
$1.05 = 1 + 0.05$
The growth factor per day is $\frac{1.05}{7}$
To go from per week to per day, use a 7th root: $(1.05)^{1/7}$. Division does not work because exponential change is multiplicative, not additive.
11. Common Mistakes
Mixing up growth factor and growth rate
These are not the same. Factor: $1.08$. Rate: $8\%$. A student might say "the growth factor is $8\%$," but the factor is $1.08$.
Forgetting the units in the output
If a function is measured in thousands, then $P(0) = 7$ means $7{,}000$, not $7$.
Using division instead of roots when changing time units
If the factor is $1.21$ every 2 months, the monthly factor is not $1.21 \div 2$. It is $\sqrt{1.21}$, because two monthly factors multiplied together must equal $1.21$.
Choosing a factor less than $1$ for growth
Increasing quantity → factor $> 1$. Decreasing quantity → factor between $0$ and $1$. This quick check eliminates many wrong answers immediately.
12. Example: Algae Population
A species of algae has a population of $2{,}300$ at the start. After 4 days, the population is $2.5$ times as large. Which model describes $A(d)$ after $d$ days?
Solution
Step 1: Initial amount
$$a = 2300$$
Step 2: Daily factor
Population is multiplied by $2.5$ over 4 days:
$$\text{daily factor} = 2.5^{1/4}$$
Step 3: Write the model
$$A(d) = 2300 \cdot \left(2.5^{1/4}\right)^d$$
Equivalently: $A(d) = 2300 \cdot (2.5)^{d/4}$
Why not $2300 \cdot 2.5^d$? Because that would mean multiplying by $2.5$ every single day, not every 4 days.
13. Real-World Model Connection
Exponential models appear in many places:
In real situations, the most important skill is not just computing values — it is understanding what the model is saying.
When a biologist says a colony doubles every 10 hours, the question is usually: What is the growth factor per hour? What formula matches those units? Which graph or statement fits?
14. Example: EV Growth
A town's electric vehicle count is modeled by $E(m) = 640 \cdot (1.18)^m$, where $m$ is in months. Find the monthly growth factor, the monthly growth rate, and the growth factor over 6 months.
Solution
Monthly growth factor
$1.18$
Monthly growth rate
$1.18 = 1 + 0.18$ → $18\%$
Growth factor over 6 months
$$(1.18)^6 \approx 2.70$$
Interpretation
In 6 months, the electric vehicle count is about $2.7$ times the starting amount.
15. Final Summary
Exponential functions model repeated multiplication, not repeated addition.
In $f(t) = a \cdot b^t$, $a$ is the initial amount and $b$ is the growth factor per time unit.
$b > 1$ → growth. $\;0 < b < 1$ → decay.
Shorter intervals use roots. Longer intervals use powers.
Always pay attention to what the exponent counts: days, hours, weeks, months.
Translate each part into words: starting value, factor, rate, and units.
A factor is not the same as a rate. $1.08$ means an $8\%$ increase.
If the output is in thousands or millions, convert carefully before making statements.
Mental check: growth → factor $> 1$, decay → factor between $0$ and $1$, shorter intervals → factor closer to $1$.