Exponentials and Logarithms

Exponentials

For any number b and positive integer n, we define exponentiation, i.e. b raised to the power n, as follows:

bn = b⋯b = b multiplied by itself n times

We can extend this definition to non-positive integers n as follows:

b^n = \begin{cases} b \cdots b & \mbox{if } n > 0 \\ 1/b^{-n} & \mbox{if } n < 0 \\ 1 & \mbox{if } n = 0 \end{cases}

For example, 23 = 2 ∙ 2 ∙ 2 = 8, 2-3 = 1/8 and 20 = 1

Exponentiation has the following properties:

b^m \cdot b^n = b^{m+n} \qquad \qquad (b^m)^n = b^{mn}

Where n > 0, we can also define b^{1/n} = \sqrt[n]{b} = the nth root of b, i.e. the number a such that a multiplied by itself n times is b. We can extend this definition to fractional exponents as follows:

b^{m/n} = \sqrt[n]{b^m} = (\sqrt[n]{b})^m where m and n are any integers.

Without getting into all the details, ba is defined for any a, and can be calculated in Excel by b^a. E.g. 3^2.1 ≈ 10.04551 and Pi()^Pi() ≈ 36.4621 where Pi() = π = 3.1415… The properties stated above for integer exponents can be extended to any exponents, namely

b^a \cdot b^c = b^{a+c} \qquad \qquad (b^a)^c = b^{ac}
b^{-a} = \frac{1}{b^a}

Logarithms

logba, called the log of a (base b) is the number c such that bc = a. Thus, the log function is the inverse of exponentiation and has the following properties:

\begin{array}{ll} \! \log_b {ac} = \log_b{a} + \log_b{c} & \qquad \log_b{a/c} = \log_b{a} - \log_b{c} \\ \! \log_b{a^c} = c \log_b{a} & \qquad \log_b{b^c} = c \end{array}

On this website, we use logs with base = 10 (called log base 10 and written simply as log a) and logs with base e where e is a special constant equal to 2.718282…. The log of a base e is called the natural log of a and is written as ln a.

4 thoughts on “Exponentials and Logarithms”

  2. Last equation on left incorrect, it should be without ‘a’ raised to ‘c’ on the right hand side of the equation (corrected latex form):

    \log_ba^c = c\log_ba

    Reply

    • Thank you very much for catching this typo. I have just corrected the mistake on the referenced webpage.
      I really appreciate your help in making the website more accurate and easier for people to use.
      Charles

      Reply

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