3.9 Applications of Logarithmic Functions

In the section on exponential functions, we explored an application of exponential functions in finance. The equation to find how much money you will have in $t$ $t$ years, $A_{t}$ $A_{t}$ , with the principal of $P$ $P$ dollars and an interest rate of $r$ $r$ is

$A_{t} = P e^{r t}$ $A_{t} = P e^{r t}$

With an understanding of logarithmic functions, we can find the interest rate or the years needed to save given the initial dollar amount and goal dollar amount. In order to solve for the interest rate $r$ $r$ ,

$\begin{matrix} A_{t} & = P e^{r t} \frac{A_{t}}{P} & = e^{r t} ln \frac{A_{t}}{P} & = r t r & = \frac{1}{t} ln \frac{A_{t}}{P} \end{matrix}$ $\begin{aligned} A_{t} & = P e^{r t} \\ \frac{A_{t}}{P} & = e^{r t} \\ \ln \frac{A_{t}}{P} & = r t \\ r & = \frac{1}{t} \ln \frac{A_{t}}{P} \end{aligned}$

Similarly, to solve for $t$ $t$ years given $r$ $r$ , $P$ $P$ , and $A_{t}$ $A_{t}$ ,

$\begin{matrix} A_{t} & = P e^{r t} \frac{A_{t}}{P} & = e^{r t} ln \frac{A_{t}}{P} & = r t t & = \frac{1}{r} ln \frac{A_{t}}{P} \end{matrix}$ $\begin{aligned} A_{t} & = P e^{r t} \\ \frac{A_{t}}{P} & = e^{r t} \\ \ln \frac{A_{t}}{P} & = r t \\ t & = \frac{1}{r} \ln \frac{A_{t}}{P} \end{aligned}$

While we will not discuss other frequencies of compounding here, we can use logarithmic functions to find $t$ $t$ years or the interest rate $r$ $r$ .

Here is an example. Suppose that you are planning to save $100 in an investment account today, and you want to pull out $200 in 10 years. What interest rate do you need in order to achieve your goal given the amount you plan to save today?

Since we solved for $r$ $r$ above, we just need to plug in the necessary variables:

$\begin{matrix} r & = \frac{1}{t} ln \frac{A_{10}}{P} = \frac{1}{10} ln \frac{200}{100} = 0.2303 \end{matrix}$ $\begin{aligned} r & = \frac{1}{t} \ln \frac{A_{10}}{P} \\ = \frac{1}{10} \ln \frac{200}{100} \\ = 0.2303 \end{aligned}$

If you invest $100 in an investment account at a 23.03% interest rate, you will have $200 in the account in 10 years. However, this does not sound very realistic—it would be extremely hard to find an investment opportunity that provides a 23.03% interest rate for 10 years.

What if the investment rate was 6%, and you wanted to know how long it takes for an initial amount of $100 to grow to $200? Again, we will plug these values into the formula we derived.

$\begin{matrix} t & = \frac{1}{r} ln \frac{A_{10}}{P} = \frac{1}{0.06} ln \frac{200}{100} = 38.38 \end{matrix}$ $\begin{aligned} t & = \frac{1}{r} \ln \frac{A_{10}}{P} \\ = \frac{1}{0.06} \ln \frac{200}{100} \\ = 38.38 \end{aligned}$

So with a 6% interest rate, your $100 investment would grow to $200 in 38.38 years.

Want to try our built-in assessments?

Use the Request Full Access button to gain access to this assessment.

Previous Next