Notes on Linear vs Exponential Growth/Decay: Price Changes, Base-e Form, and Logarithms

Linear vs. Exponential Price Change

Initial setup: price starts at $80 and is reduced by a fixed amount each day in a simple example.
- Day 0 (initial): $80
- Day 1: $80 - $4 = $76
- Day 2: $76 - $4 = $72
- Day 3: $72 - $4 = $68 (continues the same pattern)
- This is a linear function in time.
Key linear model (one-step-per-time):
- Slope m is the amount changed per day. Here, answering by points: subtracting 4 each day gives a slope of $m = -4$ .
- Intercept (y-axis, at t = 0): $b = 80$ .
- Linear function form: $f(t) = mt + b = -4t + 80 = 80 - 4t$ .
- Time variable: $t$ is the number of days elapsed (an integer in the table, but can be any real number for a model).
Summary of linear model implications:
- Change per day is additive: you add/subtract a constant each time step.
- The slope is constant; the graph is a straight line.
Transition to exponential thinking: when a price is reduced by a percentage each day, the change is multiplicative, not additive.

Exponential Price Change: 5% Reduction Per Day

Scenario: price is reduced by 5% each day, i.e., kept at 95% of the price from the previous day.
- 5% reduction means the new price is 95% of the old price.
Exponential model (discrete growth/decay):
- General form: $f(t) = q_0 \, a^t$ where a > 0 and $t o ext{time steps (days)}$ .
- Here, initial quantity (price) $q_0 = 80$ and daily factor $a = 0.95$ (since price is reduced by 5% each day).
- Explicit form: $f(t) = 80 \, (0.95)^t$
Consequences of the exponential model:
- Day 1: $f(1) = 80 \cdot 0.95 = 76$
- Day 2: $f(2) = 80 \cdot (0.95)^2 = 72.2$
- Demonstrates that at each step you multiply by the same factor, not add/subtract.
Key takeaway: linear vs exponential
- Linear: add/subtract a constant each day (slope is the constant change).
- Exponential: multiply by a constant factor each day (base a; a > 0 and a ≠ 1).
- If the table shows repeated multiplication, the model is exponential.
Notation and parameters:
- Initial quantity: $q_0 = 80$ or simply the value at time t = 0.
- Base (multiplicative factor): $a = 0.95$ .
- Discrete growth rate: $r = a - 1 = -0.05$ , so the per-step percent change is
  $r = -0.05 = -5\%.$
- Relationship to the linear form: this model is not linear in $t$ ; it is exponential in $t$ .
Base-1 plus rate intuition:
- Sometimes it helps to write the multiplicative factor as $a = 1 + r$ , so for a 5% decrease, $r = -0.05$ and $a = 0.95$ .
Continuous vs discrete growth rate quick reference:
- Discrete growth rate: $f(t) = q_0 (1 + r)^t$ with $r$ the per-step (per-day) rate.
- Continuous growth rate: transform to $f(t) = q_0 \, e^{k t}$ where $k$ is the continuous rate and satisfies $a = e^{k}$ .

From Discrete to Continuous: The Base-E Form

Alternative exponential form using base $e$ (continuous growth):
- $f(t) = q_0 \, e^{k t}$ where $k$ is the continuous growth rate.
- Relationship to the discrete base: $a = e^{k}$ and hence $k = \ln(a)$ .
Example from the 5% decay case:
- If you wanted to express $f(t) = 80 \, (0.95)^t$ in base $e$ :
- Write as $f(t) = 80 \, e^{k t}$ with $e^{k} = 0.95$ , so
  $k = \ln(0.95) \approx -0.051293.$
- Note: this k (continuous rate) is not the same as the discrete rate $r = -0.05$ , though they are related via $a = e^{k}$ .
Why use base $e$ ?
- The derivative of $e^x$ is itself: $\frac{d}{dx} e^x = e^x$ , which makes calculus neater.
- In many contexts, using the base $e$ simplifies analysis and differentiation.
Conversion rule (brief recap):
- Any exponential can be written as $q0 a^t = q0 (e^{k})^t = q_0 e^{k t}$ with $a = e^{k}$ and vice versa.
- Therefore, the continuous growth rate is $k = \ln(a)$ .

A Two-Point Method for Exponential Functions (when the initial amount is not given)

Suppose you know two later values, e.g., $f(5) = 75.94$ and $f(7) = 170.86$ , and you know the model is exponential $f(t) = q_0 \cdot a^t$ .
Set up two equations:
- $75.94 = q_0 \cdot a^{5}$
- $170.86 = q_0 \cdot a^{7}$
Eliminate $q_0$ by division:
- $\frac{f(7)}{f(5)} = \frac{q0 a^{7}}{q0 a^{5}} = a^{2}$
- Therefore $a = \sqrt{\frac{f(7)}{f(5)}} = \sqrt{\frac{170.86}{75.94}} \approx 1.5$ .
Solve for the initial quantity $q_0$ :
- From $75.94 = q0 a^{5}$ , so $q0 = \frac{75.94}{a^{5}} \approx \frac{75.94}{1.5^{5}} \approx 10.0.$
Resulting model:
- $f(t) = 10 \cdot (1.5)^t$ (with time measured in days).
- Check: $f(5) \approx 10 \cdot 1.5^{5} = 75.94$ and $f(7) \approx 10 \cdot 1.5^{7} = 170.86.$
Growth rates in this example:
- Discrete growth rate: $r = a - 1 = 0.5$ , i.e., 50% increase per day.
- Continuous growth rate: $k = \ln(a) = \ln(1.5) \approx 0.405465.$ (Note: not the same as the discrete rate.)
Important distinction clarified in this context:
- The problem can give you two later points to deduce both the base and the initial amount when the initial amount isn’t explicit.
- Once you have $a$ and $q0$ , you have the full model: $f(t) = q0 a^{t}$ .

Quick Reference: Key Formulas and Concepts

Linear model (constant change per time step):
- $f(t) = m t + b$ , with slope $m = \frac{\Delta y}{\Delta x}$ and intercept $b = f(0)$ .
- Example here: $f(t) = -4t + 80$ , with initial value $f(0) = 80$ and slope $m = -4$ .
Exponential model (per-step multiplicative change):
- $f(t) = q_0 \, a^{t}$ , where a > 0 and often written as $a = 1 + r$ .
- Discrete growth/decay rate: $r = a - 1$ .
- Example: 5% daily decrease: $f(t) = 80 (0.95)^t$ with $r = -0.05$ and $a = 0.95$ .
Base e form (continuous rate):
- $f(t) = q_0 \, e^{k t}$ with $a = e^{k}$ and $k = \ln(a)$ .
- Example: If $a = 1.5$ , then $k = \ln(1.5) \approx 0.405465$ and the discrete rate is $r = 0.5$ .
Converting between forms:
- $q0 a^{t} = q0 e^{k t}$ where $a = e^{k}$ and $k = \ln(a)$ .
Natural logarithm basics:
- $\ln x$ is the inverse of $e^{x}$ .
- If $e^{k} = a$ , then $k = \ln a$ .
- Example: $\frac{1}{\sqrt{e}} = e^{-1/2}$ and $\ln\left(e^{-1/2}\right) = -\tfrac{1}{2}.$
Properties and notes:
- The base of an exponential function must be positive and not equal to 1.
- In practice, using base $e$ often simplifies differentiation and certain analyses.
- The number $e \approx 2.718281828…$ is irrational and has many convenient properties in calculus and growth models.
Practical takeaway for exam problems:
- Identify whether changes are additive (linear) or multiplicative (exponential).
- Use the discrete form $f(t) = q_0 a^{t}$ for per-step changes; compute $a$ via direct data or via ratios when two points are given.
- If asked for continuous growth rate, convert to base $e$ form and compute $k = \ln(a)$ ; remember $a = e^{k}$ .
- Practice converting between discrete and continuous representations and interpreting the meaning of the growth rates in context.