A calculus of the absurd

9.2 Logarithms

These seem scary at first, but they’re not actually too bad.

A logarithm has a "base", and a "power". When \(\log _a(b)\) is written, it means "what needs to be raised to the power of \(a\) to get \(b\)?" For example, \(log_2(8)=3\), as \(2^3=8\).

The definition of a logarithm is that \(z=\log _b(w)\) if and only if \(w=b^{z}\). From here, we can prove a bunch of facts about the logarithm function.

For example, if we let \(z=\log _b(w)\) and \(p=\log _b(q)\) then we can then express \(\log (wq)\) in terms of \(z\) and \(p\).

\begin{equation*} \log (wq) = \log (b^z b^p) \end{equation*}

We can then use one of the law of powers, that \(b^{x}b^{y} = b^{x + y}\) 6262 This is explored above. to write that

\begin{equation*} \log (b^z b^p) = \log (b^{z+p}) \end{equation*}

After this, we can use the definition of the \(\log \) function to simplify the right-hand side of the previous equation.

\begin{equation*} \log (b^{z+p}) = z + p \end{equation*}

And from our earlier definitions of \(z=\log _b(w)\) and \(p=\log _b(q)\) we can say that 6363 This is particularly powerful because it means that we can write any multiplication as a sum (and there’s a lot more algebra that can be applied to sums than products).

\begin{align} \label {product to sum log law} \log (wq) &= z + p \notag \\ &= \log _b(w) + \log _b(q) \end{align}