Chapter 10 Logarithmic Functions
We have used logarithms to solve exponential equations. In this chapter, we consider logarithmic functions as models in their own right. We also study another base for exponential and logarithmic functions, the natural base \(e\text{,}\) which is the most useful base for many scientific applications.
In 1885, the German philosopher Hermann Ebbinghaus conducted one of the first experiments on memory, using himself as a subject. He memorized lists of nonsense syllables and then tested his memory of the syllables at intervals ranging from 20 minutes to 31 days. After one hour, he remembered less than 50% of the items, but he found that the rate of forgetting leveled off over time. He modeled his data by the function
Time elapsed  Percent remembered 
\(20\) minutes  \(58.2\%\) 
\(1\) hour  \(44.2\%\) 
\(9\) hours  \(35.8\%\) 
\(1\) day  \(33.7\%\) 
\(2\) days  \(27.8\%\) 
\(6\) days  \(25.4\%\) 
\(31\) days  \(21.1\%\) 
Ebbinghaus's model uses a logarithmic function. The graph of the data, shown above with \(t\) in minutes and \(y\) in percents, is called the "forgetting curve." Ebbinghaus's work, including his application of the scientific method to his research, provides part of the foundation of modern psychology.
Investigation 10.0.1. Interest Compounded Continuously.
We learned in SectionÂ 7.4 that the amount, \(A\) (principal plus interest), accumulated in an account with interest compounded \(n\) times annually is
where \(P\) is the principal invested, \(r\) is the interest rate, and \(t\) is the time period, in years.

Suppose you keep \(\$1000\) in an account that pays 8% interest. How much is the amount \(A\) after 1 year if the interest is compounded twice a year? Four times a year?
\begin{gather*} n = \alert{2}: A = 1000\left(1 + \frac{0.08}{\alert{2}} \right)^{\alert{2}(1)}= \\ n = \alert{4}: A = 1000\left(1 + \frac{0.08}{\alert{4}} \right)^{\alert{4}(1)}= \end{gather*} 
What happens to \(A\) as we increase \(n\text{,}\) the number of compounding periods per year? Fill in the table showing the amount in the account for different values of \(n\text{.}\)
\(n\) \(A\) \(1\) (annually) \(1080\) \(2\) (semiannually) \(\) \(4\) (quarterly) \(\) \(6\) (bimonthly) \(\) \(12\) (monthly) \(\) \(365\) (daily) \(\) \(1000\) \(\) \(10,000\) \(\) Plot the values in the table from \(n = 1\) to \(n = 12\text{,}\) and connect them with a smooth curve. Describe the curve: What is happening to the value of \(A\text{?}\)

In part (2), as you increased the value of \(n\text{,}\) the other parameters in the formula stayed the same. In other words, \(A\) is a function of \(n\text{,}\) given by \(A = 1000 (1 + \dfrac{0.08}{n} )^n\text{.}\) Use your calculator to graph A on successively larger intervals:
\(\displaystyle \text{Xmin} = 0, \text{Xmax} = 12; \text{Ymin} = 1080, \text{Ymax} = 1084\)
\(\displaystyle \text{Xmin} = 0, \text{Xmax} = 50; \text{Ymin} = 1080, \text{Ymax} = 1084\)
\(\displaystyle \text{Xmin} = 0, \text{Xmax} = 365; \text{Ymin} = 1080, \text{Ymax} = 1084\)
Use the Trace feature or the Table feature to evaluate \(A\) for very large values of \(n\text{.}\) Rounded to the nearest penny, what is the largest value of \(A\) that you can find?
As \(n\) increases, the values of \(A\) approach a limiting value. Although \(A\) continues to increase, it does so by smaller and smaller increments and will never exceed $1083.29. When the number of compounding periods increases without bound, we call the limiting result continuous compounding.
Is there an easier way to compute \(A\) under continues compounding? Yes! Compute \(1000e^{0.08}\) on your calculator. (Press 2nd LN to enter \(e^x\text{.}\)) Compare the value to your answer in part (5) for the limiting value. The number \(e\) is called the natural base. We'll compute its value shortly.

Repeat your calculations for two other interest rates, 15% and (an extremely unrealistic) 100%, again for an investment of $1000 for 1 year. In each case, compare the limiting value of \(A\text{,}\) and compare to the value of \(1000e^r\text{.}\)
\(r=0.15\) \(n\) \(A\) \(1\) \(115\) \(2\) \(\hphantom{00000}\) \(4\) \(\) \(6\) \(\) \(12\) \(\) \(3652\) \(\) \(1000\) \(\) \(10,000\) \(\) \(~~1000e^{0.15}= \) \(r=1\) \(n\) \(A\) \(1\) \(200\) \(2\) \(\hphantom{00000}\) \(4\) \(\) \(6\) \(\) \(12\) \(\) \(3652\) \(\) \(1000\) \(\) \(10,000\) \(\) \(~~1000e^{1}= \)
In part (8b), you have computed an approximation for \(1000e\text{.}\) What is the value of \(e\text{,}\) rounded to 5 decimal places?

Complete the table of values. What does \(\left(1 + \dfrac{1}{n} \right)^n\) appear to approach as \(n\) increases?
\(n\) \(100\) \(1000\) \(10,000\) \(100,000\) \(\left(1+\frac{1}{n} \right)^n \)