Exponential functions are closely related to geometric sequences. That is, sequences of the form

in which the ratio of any term to its preceding term is a non-zero constant.

An exponential function can be used to model a wide variety of real-life situations; from financial growth of investments to depreciation in the value of a vehicle; from population growth to radioactive decay.

In This Module

• We will explore various applications of the exponential function and solve related problems.

An unstable element Fermium-\(253\) (\(^{253}\)Fm) has a half-life of \(3\) days.

The mass \( M \), in milligrams, of a given amount of \(^{253}\)Fm remaining after \( t \) days is modelled by the given table.

Determine the equation of a function that models this relationship.

Solution

Half-life is the time required for a quantity to decrease to half its original amount.

Half-life is used to measure decay when working with radioactive substances in nuclear physics and chemistry.

An unstable element Fermium-\(253\) (\(^{253}\)Fm) has a half-life of \(3\) days.

The mass \( M \), in milligrams, of a given amount of \(^{253}\)Fm remaining after \( t \) days is modelled by the given table.

Determine the equation of a function that models this relationship.

Solution

If we consider the graph associated with the given table, the function to model this data appears to be a decreasing exponential function (with the value of the base between \(0\) and \(1\)).

An unstable element Fermium-\(253\) (\(^{253}\)Fm) has a half-life of \(3\) days.

The mass \( M \), in milligrams, of a given amount of \(^{253}\)Fm remaining after \( t \) days is modelled by the given table.

Determine the equation of a function that models this relationship.

Solution

When working with polynomial functions, we first used finite differences to help us identify a relationship, given a table of values.

It is apparent from the finite differences table that this data cannot be modelled by a polynomial function. A constant \( n^{\text{th}} \) difference cannot be obtained.

The \(1^{\text{st}}\) differences are not constant, but are related by a factor of \( \frac{1}{2} \); for instance, \( -250 \) is \( \frac{1}{2} \) of \( -500 \).

This same relationship occurs with the \(2^{\text{nd}}\) differences and the values of the mass, \( M \).

An unstable element Fermium-\(253\) (\(^{253}\)Fm) has a half-life of \(3\) days.

The mass \( M \), in milligrams, of a given amount of \(^{253}\)Fm remaining after \( t \) days is modelled by the given table.

Determine the equation of a function that models this relationship.

Solution

In this relationship, the ratio of consecutive values of \( M \) in the table is constant.

That is, \(\dfrac{M(3)}{M(0)}=\dfrac{M(6)}{M(3)}=\dfrac{M(9)}{M(6)}\) and so on.

This will happen for exponential behaviour when the change in the independent variable is constant. In this case, \( \Delta{t} = 3 \).

An unstable element Fermium-\(253\) (\(^{253}\)Fm) has a half-life of \(3\) days.

The mass \( M \), in milligrams, of a given amount of \(^{253}\)Fm remaining after \( t \) days is modelled by the given table.

Determine the equation of a function that models this relationship.

Solution

This common ratio of \( \frac{1}{2} \) is also present within the set of first differences and second differences. The occurrence of a common ratio is a way of identifying exponential behaviour.

An unstable element Fermium-\(253\) (\(^{253}\)Fm) has a half-life of \(3\) days.

The mass \( M \), in milligrams, of a given amount of \(^{253}\)Fm remaining after \( t \) days is modelled by the given table.

Determine the equation of a function that models this relationship.

Solution

The common ratio \( \frac{1}{2} \) provides us with the base of the exponential function used to model this data.

To determine the equation we begin with the general form of the exponential function having base \( \frac{1}{2} \).

The horizontal asymptote in this stuation is \( M = 0 \) since \( M \rightarrow 0 \) as \( t \rightarrow \infty \).

There is no vertical translation of the function \( y = \left( \frac{1}{2} \right)^x \) and so \( \textcolor{Violet}{k} = 0 \).

A horizontal translation will not affect the shape of the exponential graph and can be converted to a vertical stretch, as demonstrated in an earlier module.

Therefore, we let \( \textcolor{Mulberry}{h} = 0 \) and consider only the vertical and horizontal stretch factors to define this function.

An unstable element Fermium-\(253\) (\(^{253}\)Fm) has a half-life of \(3\) days.

The mass \( M \), in milligrams, of a given amount of \(^{253}\)Fm remaining after \( t \) days is modelled by the given table.

Determine the equation of a function that models this relationship.

Solution

The equation to model this data can thus be determined using \( M(t) = a \left( \frac{1}{2} \right)^{bt} \) where \( t \ge 0 \) is the time in days.

The initial mass of the substance, denoted \( M_0 \), is \( 1000 \text{ mg} \), so \( M(0) = 1000 \).

Using \( M(3) = 500 \), we can solve for \( b \).

Thus, \( M(t) = 1000\left(\frac{1}{2}\right)^{\frac{t}{3}}, t \ge 0 \).