LN

Returns the natural logarithm of a number.

Syntax:

LN(number)

returns the natural logarithm (the logarithm to base e) of number, that is the power of e necessary to equal number. The mathematical constant e is approximately 2.71828182845904.

Example:

LN(3)

returns the natural logarithm of 3 (approximately 1.0986).

LN(EXP(321))

returns 321.

Application:

Continuous Compound Interest

Let's say you have an initial investment of $1,000 in a savings account that offers a 5% interest rate compounded continuously. You want to know how many years it will take for your investment to double to $2,000.

The formula for continuous compound interest is: A=Pert

Where:

A is the final amount ($2,000 in this case).
P is the principal or initial amount ($1,000).
e is Euler's number (approximately 2.71828).
r is the annual interest rate (0.05).
t is the time in years (what we want to find).

To solve for t, we can use the natural logarithm.

Set up the equation: 2000=1000e0.05t
Isolate the exponential term: Divide both sides by 1000: 2=e0.05t
Use the natural logarithm to solve for the exponent: Take the natural logarithm (ln) of both sides. The key property of logarithms is that ln(ex)=x. ln(2)=ln(e0.05t) ln(2)=0.05t
Solve for t: t=ln(2)/0.05 t≈0.693/0.05 t≈13.86 years

It will take approximately 13.86 years for your initial investment to double.

Table of Growth Over Time

This table shows how the investment grows and the corresponding natural logarithm of the ratio of the final amount to the initial amount.

	Time (t in years)	Amount (A)	A/P ($A/1000)	LN(A/P)
	A	B	C	D
1	0	$1,000.00	1	0
2	1	$1,051.27	1.051	0.049742092
3	5	$1,284.03	1.284	0.249980205
4	10	$1,648.72	1.649	0.500169044
5	13.86	$2,000.00	2	0.693147181
6	15	$2,117.00	2.117	0.749999992
7	20	$2,718.28	2.718	0.999896316

As you can see from the table, the value of ln(A/P) is directly proportional to the time t. This is because the equation ln(A/P)=rt holds true. The natural logarithm allows us to linearize the exponential growth relationship, making it easy to solve for time.