# Exponential decay

In the different areas of practical knowledge, such as biology, economics, chemistry, physics, and the social sciences, we find examples of exponential decay. Where there is a decrease in some object under study, which is not linear and has a totally exponential behavior, it is generally represented with the following function:

y (x) = kax

Where k is a real number it is already positive but different from one. When at <1 or x <0, the curve is strictly decreasing.

## Definition

Exponential decay is the decrease of a quantity at a rate proportional to its own value. If a quantity N decreases at a rate proportional to the quantity present at time t, it can be written as:

N (t) = N0 and-kt

Whereis the value of N at t = 0 and k is a constant called the exponential decay function. The expression above tells us that the value of N decreases as t increases.

## Exponential decay formula

We know that mathematically:

Rate of change N (t) = - (constant) · N (t)

or

dN / dt = - kN

dN (t) / N (t) = - k dt

Integrating the previous expression:

ln N (t) = - k · t + C (1)

Where C is the integration constant.

Expression (1) can also be written as:

N (t) = e-kt andC

Considering that t = 0, N (0) = N0, so:

N (t) = N0 and-kt

Where0 is the initial value of the quantity N.

Consequently, the formula for exponential decay can be written as:

N (t) = N0 and-kt

The logical analysis of problems in the form of exponential decay is based on the product of the initial quantity of the object under study, due to its loss rate raised to one power.

Example: You have a radioactive substance that decays, at a rate of 3.5 % per hour. What percentage of the substance remains after 6 hours?

Let's first analyze what the problem tells us.

• When the substance has not yet decayed, that is, no time has elapsed, we have the 100 %:

How much t = 0 → We have 100% substance → N0 = 100 % of the substance

• What happens after an hour? We know that the substance decays at 3.5 % per hour, 3.5 % has been “lost”, that is:

100 % - 3.5 % = 96.5 %

Thus,

How much t = 1 → We have 96.5 % of substance → N = 96.5 %.

Now, in order to apply the exponential decay formula, we must know the value of k. Therefore, we write it in logarithmic form:

ln N (t) = ln (N0 and-kt)

ln [N (t) / N0) = - kt

Clearing k:

k = - ln [N (t) / N0) / t

Considering that N = 96.5 %; N0 = 100 % and t = 1:

k = 0.0356 per hour.

That is, the decay is 0.356 per hour.

Now, we can find the amount of substance we have after six hours:

N (t) = N0 and-kt

N (t) = 100 e– (0,356)(7) = 80,7 %

#### Another way to solve the problem:

When the substance is "new", that is, no time has elapsed, the substance is at 100 %. After an hour it tells us that it has lost 3.5 % of what it had at the beginning, that is, now it has:

100 % - 3.5 % = 96.5 %

Or we could also say:

96.5 % = (0.965) · (100)

In the second hour we will lose 0.965 again, but of what we had the first hour (0.965) · (100); then in the second hour we will have:

(0.965) · (0.965) · (100) = 93.1 %

In the third hour we will lose 0.965 of what we had of substance after two hours (0.965) · (0.965) · (100), that is, we will have:

(0.965) · (0965) · (0.965) · (100) = 89.8 %

We can see that every hour that passes, 0.965 of the substance that remains from the previous hour is lost. In this way, we could express a general form of the decay of the substance under study. We will call the substance S and it changes as a function of time (t):

S (t) = 100 · (0.965)t

Having the general expression for this problem, we will be able to know how much substance will be at any moment. Since the problem asks us what percentage there will be after 6 hours, we will substitute t = 6:

S (6) = 100 · (0.965)6

S (6) = 80.7 %

So, after 6 hours we have 80.7 % of the initial substance.