Natural Base: Definition & Overview

Instructor: Jennifer Beddoe

Jennifer has an MS in Chemistry and a BS in Biological Sciences.

The natural base is the base of a natural logarithm and is represented by the letter 'e'. This lesson will define natural base, give some examples of how it is used and conclude with a quiz to test your understanding.

Natural Base

The natural base, e, is sometimes referred to as the natural exponent. It is the base for the natural log (ln), which can be written as log_e(x).

e is a constant whose approximate value is 2.71828. It is used extensively in calculating growth and decay problems. It is sometimes called Euler's number after Leonhard Euler, a Swiss mathematician in the 1700s.

Along with 0, 1, π and I, e is one of the five major mathematical numbers. All of them play important and recurring roles in many areas of math, science and other fields of study.

Properties of the Natural Base

There are certain general properties that apply to the natural base.

  • e^(a + b) = (e^a) + (e^b)
  • e^(a/b) = e^a/e^b
  • e^(a^b) = e^(a*b)

e^0 = 1

e^(-a) = 1/e^a

ln(e) = 1

ln(e^x) = x

Growth and Decay

The natural base is used most frequently in calculating exponential growth and decay, whether it be in bacteria, the breakdown of carbon for carbon dating or financials such as investments or stock market gains and losses.

The general equation for growth and decay problems is A = Pe^(rt)

Where:

A = ending amount

P = initial amount

e = natural base = 2.71828

r = rate (this number will be negative for decay problems and positive for growth problems.)

t = time

Here are some examples of how this equation works.

1.) Radium-226, a common isotope of radium, has a half-life of 1620 years and a decay rate of 0.00043. Professor Johnson has a 120-gram sample of radium-226 in his laboratory. How many grams of radium-226 will remain after 200 years?

The first thing we notice about this problem is that it is a decay problem. The equation we use will not change, but the rate will be negative.

When we substitute the values we know into the equation we get:

A = 120*e^(-0.00043*200)

Then simplify

A = 120*e^(-0.086)

A = (120)*(0.91759)

A = 110 grams of the radium-226 will remain.

2.) A certain strain of bacteria is growing on your kitchen counter at a rate of 0.1386 bacteria per minute. Assuming that you start with only one bacterium, how many bacteria could be present after 60 minutes?

This is a growth problem, so the rate will be positive.

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