Decay Formula
Before going to learn the decay formula, let us know what is decay (or) exponential decay. "Decay" means "decrease". If the rate of decrease of a quantity is proportional to its current value, then we say that it is subject to exponential decay. Let us assume the initial value of a quantity is \(P_0\) and its current value is \(P\). Its rate of change is given by the derivative \(\dfrac{dP}{dt}\). Minus sign is because it is exponential decay. By the definition of exponential decay,
\( \begin{align} \dfrac{dP}{dt} &\propto P\\[0.2cm] \dfrac{dP}{dt} &= kP \end{align}\)
Here, k is the constant of proportionality.
Let us understand the decay formula using solved examples in the following sections.
What Is Decay Formula?
There are multiple formulas available for dealing with exponential decay problems depending on the available information. One of them can be derived by using the above differential equation.
\( \begin{align} \dfrac{dP}{dt} &= kP\\[0.2cm] \dfrac{dP}{P} &= kdt\\[0.2cm] \text{Taking }& \text{integral on both sides},\\[0.2cm] \ln P &= kt+C_1\\[0.2cm] P&=e^{kt+C_1}\\[0.2cm] P &= e^{kt}\cdot e^{C_1}\\[0.2cm] P&=Ce^{kt} \,\,\,\, [\because e^{C_1}= \text{a constant} = C] \end{align}\)
We know that the initial value of \(P\) is \(P_0\).
Thus, \(P_0 = Ce^0 \Rightarrow C=P_0\).
Substituting this in the above equation, \(P=P_0e^{kt}\).
This is one of the decay formulas. We have other decay formulas as well.
\( \begin{align} f(x)&=ab^x\\[0.2cm] f(t)&=a(1r)^t\\[0.2cm] P&=P_0e^{kt} \end{align}\)
In these formulas,
a (or) P\(_0\) = Initial amount
r = Rate of decay
k = constant of proportionality
x (or) t = time (time can be in years, days, (or) months, whatever you are using should be consistent throughout the problem).
In exponential decay, always 0 < b < 1.
Solved Examples Using Decay Formula

Example 1
David bought a new truck for $50,000. The value of the truck decreases exponentially (depreciates) at a rate of 13% per year. Then what is the value of the truck after 10 years? Round your answer to two decimals.
Solution:
The initial value of the truck is, a = $50,000.
The rate of decay is, r = 13% = 0.13.
The time is t = 10 years.
Using the exponential decay formula:
f(t) = a (1  r)^{t}
f(10) = 50000 (1  0.13)^{10}^{ }= 12421.17
Answer: The value of the truck after 10 years = $12,421.17.

Example 2:
The halflife of a certain element is 4000 years. Find its exponential decay model rounding the proportionality constant to 4 decimals.
Solution:
Using the given data, we can say that the given element is decaying and hence we use the formula of exponential decay.
\(P=P_0e^{kt}\)
Here, \(P_0\) = initial amount of the element.
It is given that its halflife is 4000 years. It means
When t = 4000, P = Half of the initial amount of the given element = \(\dfrac{P_0}{2}\).
Substitute all these values in the formula of exponential decay:
\(\dfrac{P_0}{2}=P_0e^{k(4000)}\)
Dividing both sides by P\(_0\),
\(0.5 = e^{ k (4000)}\)
Taking "ln" on both sides,
\(\ln 0.5 = 4000k\)
Dividing both sides by 4000,
\(k = \dfrac{\ln 0.5}{4000} ≈ 0.00017329\)
Thus, the exponential decay model of the given element is,
\(P=P_0e^{0.00017329t}\)
Answer: \(P=P_0e^{0.00017329t}\)