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.