Another property of the logarithm is that it is multivalued.

Consider: `e`^{0} = 1, and `e`^{2πi} = 1 (where `e` is Euler's number and `i` is iota). So ln(1) can be 0, or it can be 2π`i` (or 4π`i`, or -2π`i`, etc.).

Now take into account that ln(`a` * `b`) = (ln(`a`) + ln(`b`)), and (`a` * 1) = `a`, so 1 can be substituted for `b` in *any* natural logarithm. For any `x` that satisfies `e`^{x} = `a`, there's also `e`^{(x + 2Zπi)} = `a` (where **Z** is the set of integers), i.e. if `x` is a solution to ln(a) then so is (`x` + 2**Z**π`i`).

But that's not all: one more property of logarithms is that:

log_{b}(`a`) = (log_{c}(`a`) / log_{c} (`b`)), for *any* `c`.

Choosing `e` for `c`, when `b` is already `e`, gives:

ln(`a`) = (ln(`a`) / ln(`e`)) = ((ln(`a`) + 2**Z**π`i`) / (1 + 2**Z**π`i`)). So if `x` is a solution to ln(`a`), so is ((`x` + 2**Z**π`i`) / (1 + 2**Z**π`i`)).

Finally, choosing `e` for `c` above, *any* logarithm to *any* base is multivalued.

log_{b}(`a`) = (ln(`a`) / ln(`b`)) = (((ln(`a`) + 2**Z**π`i`) / (1 + 2**Z**π`i`)) / ((ln(`b`) + 2**Z**π`i`) / (1 + 2**Z**π`i`)))

That's a countable infinity of possible answers, with four dimensions (or degrees of freedom).