infinite-logarithms-paradox

Infinite Logarithms of Every Number

Summary: §516. Every number has infinitely many logarithms, only one of which is real. From $-1=1/(-1)$ we get $\log (-1)=-\log (-1)$ but $\log (-1)\ne 0$; combining with $\log (-1{)}^2=\log 1=0$ produces $2\log (-1)=0$, an apparent contradiction unless logarithms are infinitely-multi-valued.

Sources: chapter21 §516.

Last updated: 2026-05-12

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The setup

Take any number $a$. Half of $a$ should be just $a/2$. But $\log$ rules give two values:

$\frac{1}{2}a=\frac{a}{2}\text{and}\frac{1}{2}a=\frac{a}{2}+\log (-1),$

since $a+2\log (-1)=a+\log (-1{)}^2=a+\log 1=a$. Hence both candidates for $\frac{1}{2}a$ satisfy “doubled gives $a$”, but only one is the real value.

Similarly, from $1^{1/3}$ being not only $1$ but also $\frac{-1\pm \sqrt{-3}}{2}$ (the cube roots of unity), one third of $a$ has three candidates:

$\frac{a}{3},\frac{a}{3}+\log \frac{-1+\sqrt{-3}}{2},\frac{a}{3}+\log \frac{-1-\sqrt{-3}}{2}.$

And so on for $1/n$: $n$ candidates of which only one is real.

$\log (-1)\ne 0$ but $2\log (-1)=0$

A key subtlety. From $-1=1/(-1)$:

$\log (-1)=\log 1-\log (-1)=-\log (-1),$

so $\log (-1)=-\log (-1)$. By the “real-number” logic this forces $\log (-1)=0$, but then $\log (-1)=\log 1$ collapses $-1=1$. Resolution: $\log (-1)$ is complex, not zero, and the “ratio of $\log (-1)$ to $i$ is finite”, giving $\log (-1)=i\pi$ in the modern reading. We have $-\log (-1)=\log (-1)$ holding in the modular sense $\log (-1)\equiv -\log (-1)(\mathrm{mod}2\pi i)$.

But $\log (-1{)}^2=\log 1=0$, hence $2\log (-1)\equiv 0$ — meaning $2\log (-1)=2k\pi i$ for some integer $k$, the simplest being $k=0$. Modern terms: this is the multi-valued $\log z$ on the Riemann sphere.

The §517 statement

Every number has an infinite number of logarithms, among which no more than one is real. (source: chapter21, §516)

The complex logarithms of unity include $2\log (-1),3\log \frac{-1+\sqrt{-3}}{2},4\log (-1),4\log (\pm i)$, “and an infinite number of others may be obtained by the extraction of roots of unity”. In modern notation, $\log 1=2k\pi i$ for all $k\in \mathbb{Z}$.

The exp-series argument (§516)

Why infinitely many logarithms? If $z=\log a$, then $a=e^z$, which expands as

$a=1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\cdots$

— an equation of infinite degree in $z$. By the principle that a degree-$n$ polynomial equation has $n$ roots (counted with complex roots), an infinite-degree equation has infinitely many. Each is a legitimate logarithm of $a$; only one is real.

This is one of Euler’s clearest pre-1748 anticipations of the modern multi-valued logarithm.

Connection to the discrete-point paradox

The §516 paradox is the algebraic shadow of the §515 discrete-points-below-asymptote geometric anomaly. The dense scatter of points below the asymptote of the logarithmic-curve corresponds — via the equation $y=a{e}^{x/b}$ — to non-principal branches of $\log$ becoming real after an even-power root extraction.

Euler’s own assessment

It is not clear how this paradox can be reconciled with the usual idea of quantity. (source: chapter21, §516)

Although we have explained the second paradox, the first one still retains its vigor, insofar as we have shown that the logarithmic curve has an infinite number of discrete points below the axis. (source: chapter21, §516)

Euler is satisfied that the multi-valued-logarithm account is consistent, but not that it dissolves the discrete-point anomaly — that remains a genuine pre-calculus puzzle.

Related pages

• chapter-21-on-transcendental-curves
• logarithmic-curve
• discrete-points-below-asymptote — the geometric face of this paradox.
• transcendental-curves — broader context.