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Exponent Addition

Prerequisites

Definition of Exponentiation | \(x^{y} = \prod_{i = 1}^{y} x\)

Description

This equation demonstrates a way to separate expressions of the form \(\htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y} + \htmlClass{sdt-0000000043}{z}}\). It can also be used in reverse to aggregate expressions of the form \(\htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y}} \cdot \htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000043}{z}}\).

\[\htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y} + \htmlClass{sdt-0000000043}{z}} = \htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y}} \cdot \htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000043}{z}}\]

Symbols Used:

This is a secondary symbol for any generic variable. It can hold any value, whether that be an integer or a real number, or a complex number, or a matrix etc.
\( x \)

This is a symbol for any generic variable. It can hold any value, whether that be an integer or a real number, or a complex number, or a matrix etc.
\( z \)

This is a tertiary symbol for any generic variable. It can hold any value, whether that be an integer or a real number, or a complex number, or a matrix etc.

Derivation

  1. Consider the Definition of Exponentiation:
    \[\htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y}} = \htmlClass{sdt-0000000097}{\prod}_{\htmlClass{sdt-0000000018}{i} = 1}^{\htmlClass{sdt-0000000017}{y}} \htmlClass{sdt-0000000003}{x}\]
    meaning that \(\htmlClass{sdt-0000000003}{x}\) is multiplied by itself \(\htmlClass{sdt-0000000017}{y}\) times.

    By substituting \(\htmlClass{sdt-0000000043}{z}\) for \(\htmlClass{sdt-0000000017}{y}\), we can also say that...
    \[\htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000043}{z}} = \htmlClass{sdt-0000000097}{\prod}_{\htmlClass{sdt-0000000018}{i} = 1}^{\htmlClass{sdt-0000000043}{z}} \htmlClass{sdt-0000000003}{x}\]
  2. From the Definition of Exponentiation, it follows that:
    \[\htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y} + \htmlClass{sdt-0000000043}{z}} = \htmlClass{sdt-0000000097}{\prod}_{\htmlClass{sdt-0000000018}{i} = 1}^{\htmlClass{sdt-0000000017}{y} + \htmlClass{sdt-0000000043}{z}} x\]
  3. We can now use the fact that Multiplication is Commutative (the order that you multiply things does not matter) to say that...
    \[\htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y} + \htmlClass{sdt-0000000043}{z}} = \htmlClass{sdt-0000000097}{\prod}_{\htmlClass{sdt-0000000018}{i} = 1}^{\htmlClass{sdt-0000000017}{y}} x \cdot \htmlClass{sdt-0000000097}{\prod}_{\htmlClass{sdt-0000000018}{i} = 1}^{\htmlClass{sdt-0000000043}{z}} x\]
  4. Notice that the products are equal to \(\htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y}}\) and \(\htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000043}{z}}\) respectively (see step (1)). We can therefore simplify to...
    \[\htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y} + \htmlClass{sdt-0000000043}{z}} = \htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000017}{y}} \cdot \htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000043}{z}}\]
    as required.

Example

What is the value of \(3^{2 + 3}\)?

We can use the equation to say that

\[3^{2 + 3} = 3^{2} \cdot 3^{3}\].

We can then calculate that \(3^2 = 3 \cdot 3 = 9\) and that \(3^3 = 3 \cdot 3 \cdot 3 = 27\)

From here it follows that...

\[3^{2+3} = 9 \cdot 27 = 243\]