Pre-Calculus Exponential and logarithmic functions (advanced)

Compound interest and the number e

Q: What is the number e?

The natural base, about 2.71828. It is the limit of one plus one over n to the n as n grows without bound, and the base of continuous growth.

Q: What is the difference between the two interest formulas?

A equals P times one plus r over n to the n t compounds n times per year; A equals P e to the r t compounds continuously, the limiting case.

Q: How do you solve for time in continuous growth?

Isolate the exponential and take the natural log: t equals the natural log of A over P, divided by r.

Q: Why does e show up in compound interest?

Because compounding more and more often drives one plus r over n to the n toward e to the r; continuous compounding is that limit.

20 practice questions 0 video lessons Theory + worked examples

The Number e and Compound Interest

Texas Precalculus (TEKS) • Standard P.2(N) • Exponential & Logarithmic Functions

The Number e and Compound Interest is a topic in Exponential & Logarithmic Functions in the Texas Essential Knowledge and Skills (Precalculus, §111.42). It is aligned to Standard P.2(N), which requires students to analyze situations modeled by exponential functions.

The natural base $e\approx 2.718$ is the base of continuous growth; money compounds by $A=P\!\left(1+\dfrac{r}{n}\right)^{nt}$ discretely and $A=Pe^{rt}$ continuously.

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Theory

Money that earns interest on its interest grows exponentially. With $P$ the principal, $r$ the annual rate, and $t$ years:

Compounded $n$ times a year: $A=P\left(1+\dfrac{r}{n}\right)^{nt}$.
Compounded continuously: $A=Pe^{rt}$.

The number $e\approx 2.71828$ is the natural base. It arises as the limit of $\left(1+\dfrac{r}{n}\right)^{n}$ as $n\to\infty$ — compounding as often as possible.

\(e\) is not arbitrary. It is the base for which the growth rate equals the current value, which makes \(e^{x}\) central to all of calculus.

Continuous growth follows $A=Pe^{rt}$.

The discrete and continuous compound-interest formulas.

The two compound-interest formulas:

\[A=P\left(1+\dfrac{r}{n}\right)^{nt},\qquad A=Pe^{rt}\]

A equals P times one plus r over n to the n t; A equals P times e to the r t

Solve for \(t\) with the natural log: from \(A=Pe^{rt}\), \(t=\dfrac{\ln(A/P)}{r}\).

How to use the compound-interest formulas

Identify $P,r,t$, and $n$ (or recognize continuous compounding).
Choose the discrete or continuous formula.
Substitute and evaluate, keeping the rate as a decimal.
For time, isolate the exponential and take $\ln$.

Example 1 — Compounded monthly

Invest $\$2000$ at $6\%$ compounded monthly. Find the value after 5 years.

Solution

Use $A=P(1+\dfrac{r}{n})^{nt}$ with $n=12$.

$A$	$=$	$2000\left(1+\dfrac{0.06}{12}\right)^{12\cdot 5}$
	$=$	$2000(1.005)^{60}\approx \$2697$

value is about 2697 dollars

Example 2 — Compounded continuously

Repeat with continuous compounding.

Solution

Use $A=Pe^{rt}$.

$A$	$=$	$2000e^{0.06\cdot 5}$
	$=$	$2000e^{0.3}\approx \$2700$

Slightly more than monthly compounding.

continuous value is about 2700 dollars

Example 3 — Solve for time

How long for $\$1000$ to double at $5\%$ compounded continuously?

Solution

Set $2000=1000e^{0.05t}$, then use the natural log.

$2$	$=$	$e^{0.05t}$
$\ln 2$	$=$	$0.05t$
$t$	$=$	$\dfrac{\ln 2}{0.05}\approx 13.9\ \text{yr}$

doubling time is about 13.9 years

Example 4 — The meaning of e

What does $A=Pe^{rt}$ represent compared with monthly or daily compounding?

Solution

As the number of compounding periods $n$ grows without bound, $\left(1+\dfrac{r}{n}\right)^{n}\to e^{r}$.

\[\lim_{n\to\infty}\left(1+\dfrac{r}{n}\right)^{nt}=e^{rt}\]

So continuous compounding is the limit of compounding ever more often.

continuous compounding is the limit of compounding more and more often

Common pitfalls

Rate as a decimal. $6\%$ is $r=0.06$, not $6$.

Match $n$ to the compounding. Monthly $n=12$, quarterly $n=4$, daily $n=365$.

Continuous uses $e$, not a large $n$. Use $A=Pe^{rt}$ directly.

Frequently asked questions

What is the number e?

The natural base, $e\approx 2.71828$. It is the limit of $\left(1+\dfrac{1}{n}\right)^{n}$ as $n\to\infty$ and the base of continuous growth.

What is the difference between the two interest formulas?

$A=P(1+r/n)^{nt}$ compounds $n$ times per year; $A=Pe^{rt}$ compounds continuously, the limiting case.

How do you solve for time in continuous growth?

Isolate the exponential and take the natural log: $t=\dfrac{\ln(A/P)}{r}$.

Why does e show up in compound interest?

Because compounding more and more often drives $\left(1+\dfrac{r}{n}\right)^{n}$ toward $e^{r}$; continuous compounding is that limit.

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\(A\)	\(=\)	\(2000\left(1+\dfrac{0.06}{12}\right)^{12\cdot 5}\)
	\(=\)	\(2000(1.005)^{60}\approx \$2697\)

\(A\)	\(=\)	\(2000e^{0.06\cdot 5}\)
	\(=\)	\(2000e^{0.3}\approx \$2700\)

\(2\)	\(=\)	\(e^{0.05t}\)
\(\ln 2\)	\(=\)	\(0.05t\)
\(t\)	\(=\)	\(\dfrac{\ln 2}{0.05}\approx 13.9\ \text{yr}\)

Compound interest and the number e

The Number e and Compound Interest

📖 Theory

How to use the compound-interest formulas

Common pitfalls

Frequently asked questions

Theory