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SectionLogarithms and Exponential Models

With the aid of logarithms, we can now gather more information from exponential functions. In this section we will discuss how to do this using various real world examples.

SubsectionAnnual to Exponential Growth Rate

Recall that the annual growth rate is the value \(b-1\) in the exponential equation

\begin{equation*} A(t)=a(b)^t. \end{equation*}

If we want to find the continuous growth rate which is equivalent to the annual growth rate above, we first rewrite the function: \begin{align*} A(t)\amp =a(b)^t\\ \amp =a(e^r)^t. \end{align*}

Next we simply need to find the value of \(r \) which keeps the above equation true. To solve for \(r \) we apply the properties of the logarithm as discussed in the previous section. Let's take a look at a specific example of this process.

Example219

Let \(A(t)=10(.75)^t\text{.}\) Write \(A(t)\) in the form \(A(t)=ae^{rt}\) and state the continous rate equivalent to the annual decay rate of \(25\% \text{.}\)

Solution

We want to find values \(a,r \) such that

\begin{equation*} 10(.75)^t=ae^{rt} \end{equation*}

for all values \(t \text{.}\) Rewriting the right hand side we get

\begin{equation*} ae^{rt}=a(e^r)^t \end{equation*}

which now is in the form \(a(b)^t \text{.}\) Since we want

\begin{equation*} 10(.75)^t=a(e^r)^t, \end{equation*}

we must have \(a=10 \text{,}\) and \(e^r=.75 \text{.}\) To find the appropriate value of \(r \) we just need to solve the equaion \(e^r=.75 \) for \(r \text{.}\) Applying natural log to each side we get \begin{align*} e^r \amp = .75\\ \ln(e^r)\amp = \ln(.75)\\ r\amp =\ln(.75) \end{align*} where this last equality follows due to property \(4 \) in the previous section. So, we can rewrite \(A(t)\) as

\begin{equation*} A(t)=10e^{\ln(.75)\cdot t}. \end{equation*}

Furthermore, since \(\ln(.75)\approx -.2876 \text{.}\) Then a continuous decay rate of about \(28.76\%\) is equivalent to an annual decay rate of \(25\%\text{.}\)

Notice that in this previous example it was not necessary for us to use the natural logarithm. In fact, you could use a logarithm with any base you desire and still get a solution. However, since natural log has base \(e \) it is the inverse of the exponential \(e^t \) thus giving us some cancelations and hence a nicer looking solution.

SubsectionDouble Time and Half Life

Another important piece of information we can now gather from exponentials is how to identify which input value corresponds to a given output value.

Example220

Suppose we are given the function

\begin{equation*} A(t)=10(1.75)^t \end{equation*}

which represents the value in dollars of an investment after \(t \) years. Identify the year in which the investment would reach $40.

Solution

We start by setting the function equal to 40:

\begin{equation*} 40=10(1.75)^t. \end{equation*}

Now, we divide each side by \(10 \) and apply a logarithm to each side. \begin{align*} 40\amp = 10(1.75)^t\\ 4\amp = (1.75)^t\\ \log(4)\amp = \log (1.75)^t\\ \log(4)\amp = t\cdot \log(1.75) \end{align*} Now, we simply divide each side by \(\log(1.75)\) to get

\begin{equation*} t=\frac{\log(4)}{\log(1.75)}\approx 2.5\ \ \text{years}. \end{equation*}
Caution221

In the solution for the last example, it is important that we first divided each side by the coefficient of the exponential (in this last example this coefficient was \(10 \)) before applying the logarithm. If we didn't do this then, more often than not, we will make a calculation mistake.

This is due to the behavior of logarithms with multiplication. In particular,

\begin{equation*} \log(10(1.75)^t)=\log(10)+\log((1.75)^t). \end{equation*}

Often this property of logarithms is forgotten, and thus leads to mistakes. Even if we remember to apply this property, this leads to much more work than is necessary.

There are a few particular input values which are solved for quite frequently. In fact, they are used often enough to have their own name.

Doubling Time

The Doubling time is the amount of time it takes for a quantity to double in value or size. If we are finding the doubling time of the exponential function

\begin{equation*} A(t)=a(b)^t \end{equation*}

where \(1\lt b \text{,}\) then we are solving the equation

\begin{equation*} 2a=a(b)^t \end{equation*}

for \(t \text{.}\)

Half Life

The Half Life is the amount of time it takes for a quantity to shrink to half its value or size. If we are finding the half life of the exponential function

\begin{equation*} A(t)=a(b)^t \end{equation*}

where \(b\lt 1 \text{,}\) then we are solving the equation

\begin{equation*} \frac{1}{2}a=a(b)^t \end{equation*}

for \(t \text{.}\)

Notice that the requirement for \(b \) to be greater than \(1 \) in doubling time and less than \(1 \) in half life is simply to have the statement make sense. For instance, we can only solve for the time it takes for an investment to double if the investment is increasing in value, and it is only increasing in value when \(1\lt b \text{.}\)

Another important thing to notice in these definitions is that we could solve for the doubling time or half life without knowing the value of the principal (or initial) value \(a \text{,}\) we need only know the rate or growth factor. For example, in the formula for doubling time we are solving for the value of \(t \) in the formula

\begin{equation*} 2a=a(b)^t. \end{equation*}

If we were to divide each side by \(a \) we would achieve the formula

\begin{equation*} 2=(b)^t. \end{equation*}

Solving this equation for \(t \) is the same as solving the original equation for \(t \) and yet this new equation does not involve \(a \) at all.