Logjam 2: Power relations
The logarithmic conversion of multiplication into addition is based on the formula:
\[b^{m}b^{n} = b^{m + n}\]
\[\begin{matrix} m \\ \overbrace{\begin{matrix} b b \cdots b \\ \end{matrix}} \\ \end{matrix} \times \begin{matrix} n \\ \overbrace{\begin{matrix} b b \cdots b \\ \end{matrix}} \\ \end{matrix} = \begin{matrix} {m + n} \\ \overbrace{\begin{matrix} b b \cdots b \\ \end{matrix}} \\ \end{matrix}\]
This says that product of the products of m and n bs is the product of (m + n) bs. It is an expression of the associativity of multiplication, for those who know what that means:
\[\left( {ab} \right)c = a\left( {bc} \right)\]
Another important formula:
\[\left( b^{m} \right)^{n} = b^{mn}\]
\[\begin{matrix} n \\ \overbrace{\begin{matrix} \begin{matrix} m \\ \overbrace{\begin{matrix} b b \cdots b \\ \end{matrix}} \\ \end{matrix} & \begin{matrix} m \\ \overbrace{\begin{matrix} b b \cdots b \\ \end{matrix}} \\ \end{matrix} & \cdots & \begin{matrix} m \\ \overbrace{\begin{matrix} b b \cdots b \\ \end{matrix}} \\ \end{matrix} \\ \end{matrix}} \\ \end{matrix} = \begin{matrix} {mn} \\ \overbrace{\begin{matrix} b b \cdots b \\ \end{matrix}} \\ \end{matrix}\]
This says that the n-th power of the m-th power of b is the product of mn bs.
So far we have only looked at whole number powers. However the last formula will lead the way to defining fractional powers, while the formula for addition of powers will lead to a meaning for negative powers. The extension to real numbers depends on “continuity” with respect to powers — this means that as fractions approach closer to a real number the power approaches or “coverges” to a single well-defined real number.
So let’s try to find a fractional power:
\[\left( b^{1/m} \right)^{m} = b^{m/m} = b^{1} = b\]
So \(b^{1/m}\) is the m-th root of b or \(\sqrt[m]{b}\). So a definition consistent with the power of a power formula is:
\[b^{n/m} = \left( b^{1/m} \right)^{n} = \left( \sqrt[m]{b} \right)^{n}\]
A similar process from the product of two powers formula gives:
\[b^{- m} = 1/b^{m}\]
[Remembering/defining that \(b^{0} = 1\).]
These considerations, along with the logarithm concept, allow us to make even more connections. Since \(b^{\prime} = b^{\log_{b}b^{\prime}}\) and \(b^{\log_{b}x} = x\):
\[b^{\log_{b}x} = x = {b^{\prime}}^{\log_{b^{\prime}}x} = \left( b^{\log_{b}b^{\prime}} \right){}^{\log_{b^{\prime}}x} = b^{\log_{b}b^{\prime}\log_{b^{\prime}}x}\]
Equating the ends and antilogging:
\[\log_{b}b^{\prime}\log_{b^{\prime}}x = \log_{b}x\]
Or more usefully:
\[\log_{b^{\prime}}x = \frac{\log_{b}x}{\log_{b}b^{\prime}}\]
This allows easy conversion from one base to another.