Growth and Decay

Compare constant differences with constant ratios

If a quantity starts at 800 and loses 80 each period, then $Q_n=800-80n$ while the model remains physically meaningful. Its successive differences are −80.

If it instead loses 20% of its current value each period, then each update retains 80%:

The first values are 800, 640, 512 and 409.6. The decreases become smaller, but the ratio of one value to the previous value stays 0.8. A decreasing graph alone does not identify either model. Examine the increments, ratios and mechanism.

For a constant per-period proportional change $r$, the growth factor is $b=1+r$. Positive growth has $b>1$; proportional decay has $0<b<1$. The base must stay positive in this ordinary real-valued model. Discrete counts may require rounding or a probabilistic interpretation, so a predicted 409.6 objects is not automatically a literal count.

Fit a growth factor from two observations

Suppose a positive quantity is 500 initially and 665.5 after three equal periods. Under the assumption of a constant growth factor,

The fitted increase is 10% per period. The positive cube root is selected because the stipulated proportional model has $b>0$.

The observations fix a factor within that chosen model. They do not prove the factor stayed constant between observations or will stay constant afterwards. More measurements, residuals and knowledge of the process are needed to assess that assumption. A fitted function and a verified description of reality are different things.

Solve a decay threshold and check the adjacent periods

When does $Q_n=800(0.8)^n$ first fall to 300 or below?

Taking logarithms gives $n\ln(0.8)\le\ln(3/8)$. Because $\ln(0.8)$ is negative, division reverses the inequality:

For observations at completed integer periods, the first qualifying period is 5. Check both sides of the threshold:

Rounding 4.3955 to the nearest integer would give the wrong stopping period. The condition requires the least integer at or above the threshold. If changes occur continuously, the fractional time may be meaningful instead; state which process is being modelled.

Interpret half-life as a repeated proportional reduction

A half-life $H>0$ is the interval over which the modelled quantity is multiplied by $1/2$. The exponential law is

For a stipulated initial mass of 100 g and half-life of five hours,

After 5, 10 and 15 hours, the values are 50 g, 25 g and 12.5 g. Half-life is not the time to reach zero. For every finite $t$, this model with $Q_0>0$ remains positive.

A physical half-life model may describe an expected quantity or a macroscopic approximation. It is not a claim that every individual item waits exactly one half-life before changing.

Distinguish a per-period rate from a continuous rate

An exponential law can also be written

where $k$ has reciprocal-time units. The product $kt$ is dimensionless. For a factor $b$ per unit of time, $k=\ln b$ per that unit. A 3% effective increase per year therefore corresponds to $k=\ln(1.03)$ per year, not $k=0.03$ per year.

The time to double under the 3% per-year factor is

A continuous-rate model with $k=0.03$ per year doubles in

These are different models. For a decay half-life $H$, the matching continuous rate is $k=-\ln2/H$. OpenStax’s exponential-function treatment supplies the underlying growth and decay framework.

Decay may apply to an excess above a baseline

Not every decaying quantity approaches zero. Let $T(t)$ denote the numerical temperature in degrees Celsius and $t$ the numerical time in hours. Consider the stipulated model

The model starts at 100 °C. Its excess above the 20 °C background halves every four hours. The total temperatures after 4, 8, 12 and 16 hours are 60, 40, 30 and 25 °C.

It would be wrong to halve the total temperature from 100 to 50 °C after four hours. The quantity obeying the simple proportional law is $T(t)-20$, not $T(t)$. The model approaches 20 °C from above and never reaches it at a finite time.

The fixed background and constant decay coefficient are assumptions. A changing environment or different transfer mechanism would require a different model. The mathematical example is not a temperature prediction for a particular object.

Combine changing factors without adding percentages

A 20% increase followed by a 20% decrease multiplies the original quantity by

The final quantity is 4% below the starting quantity. The percentage changes use different bases. To recover after a 20% loss, the required increase on the remaining amount is $1/0.8-1=0.25$, or 25%.

With different factors $b_1,b_2,\ldots,b_n$ and no other additions or removals, use $Q_n=Q_0b_1b_2\cdots b_n$. Equal additions, harvesting, repayments or replenishment cannot generally be hidden inside one unchanged exponential factor. They require an updated balance equation.

Keep the model’s scope visible

Before using a result, identify the measured quantity, time unit, initial value, applicable interval and update mechanism. Decide whether the factor concerns the whole quantity or only a component above a baseline.

A population cannot grow exponentially forever in a fixed-resource setting. A depreciation curve is a mathematical convention, not a guaranteed sale price. A financial nominal growth rate does not by itself describe purchasing power. These are reasons to define the model’s domain and intended use, not reasons to avoid modelling.

Related topics

Use Mathematical Modelling to assess assumptions and residuals. Compound Interest handles dated deposits, while Loans and Amortization models periodic withdrawals that reduce a debt balance.