\(\pi^e\) or \(e^\pi\)?

The hook

Which is bigger, \(\pi^e\) or \(e^\pi\)? Your calculator will tell you in a second: \(e^\pi \approx 23.14\) and \(\pi^e \approx 22.46\). But a calculator only tells you that; it does not tell you why, and it cannot help with infinitely many similar questions at once. Is \(3^\pi\) bigger than \(\pi^3\)? Are there any whole numbers with \(n^m = m^n\) apart from the obvious \(n = m\)?

All these questions turn out to be about a single curve, \(y = \ln x\), and the straight lines that cross it. This question shows how a sketch, used carefully, can settle comparisons that look hopeless by algebra, and how the number \(e\) emerges as the natural dividing point.

Before you start

\(b\ln a = a\ln b\) is the same as \[\frac{\ln a}{a} = \frac{\ln b}{b}.\] So every comparison of \(a^b\) with \(b^a\) is a comparison of the values of the single function \(h(x) = \frac{\ln x}{x}\) at \(a\) and \(b\). And \(\frac{\ln a}{a}\) is the gradient of the line from the origin to the point \((a, \ln a)\) on the curve \(y = \ln x\). That is why the question is about lines.

Working it through

Stage 1 The tangent through the origin

The line and curve must meet, \(mx_0 = \ln x_0\), and have the same gradient, \(m = 1/x_0\). Substituting the second into the first, \(1 = \ln x_0\), so \(x_0 = e\) and \(m = 1/e\). The tangent from the origin touches the curve at \((e, 1)\).

If instead \(y = mx\) crosses the curve at \(x = a\) and \(x = b\), then \(\ln a = ma\) and \(\ln b = mb\), so \(\frac{\ln a}{a} = m = \frac{\ln b}{b}\), which gives \(b\ln a = a\ln b\) and hence \(a^b = b^a\).■

For the sketch: a line through the origin that is slightly less steep than the tangent cuts the curve twice, once on each side of the point of tangency, so \(a < e < b\). Lines steeper than \(1/e\) miss the curve altogether, because \(\ln x\) is concave and lies entirely below its tangent \(y = x/e\).

Stage 2 Lines that cut the \(y\)-axis above the origin

From \(\ln p = mp + c\) and \(\ln q = mq + c\), \[\frac{\ln p}{p} = m + \frac{c}{p}, \qquad \frac{\ln q}{q} = m + \frac{c}{q}.\] Since \(c > 0\) and \(p < q\), \(\frac{c}{p} > \frac{c}{q}\), so \(\frac{\ln p}{p} > \frac{\ln q}{q}\). Multiplying by \(pq > 0\) gives \(q\ln p > p\ln q\), so \(p^q > q^p\).■

Geometrically: from the origin, the point \((p, \ln p)\) is seen at a steeper angle than \((q, \ln q)\) when the line through them crosses the \(y\)-axis above the origin.

Stage 3 Chords to the right of \(e\)

Because \(\ln x\) is concave, its gradient \(1/x\) decreases, so the chord from \(p\) to \(q\) is less steep than the tangent at \(p\): its gradient \(s\) satisfies \(s < \frac1p\). The chord’s \(y\)-intercept is \[c = \ln p - sp > \ln p - \frac{1}{p} \cdot p = \ln p - 1 \geq 0,\] using \(p \geq e\). So the chord crosses the \(y\)-axis above the origin.■

Take \(p = e\) and \(q = \pi\). The chord between them has a positive intercept, so by Stage 2, \(p^q > q^p\): that is, \[e^\pi > \pi^e.\]

Stage 4 Chords parallel to the tangent

Now suppose the chord from \(p\) to \(q\) has gradient exactly \(\frac1e\), the same as the tangent \(y = \frac{x}{e}\).

The chord is the line \(y = \frac{x}{e} + c\) for some \(c\). Its endpoints lie on the curve, which lies on or below the tangent \(y = \frac{x}{e}\), and touches it only at \(x = e\). The endpoints cannot both be at \(x = e\), so at least one lies strictly below the tangent, and therefore the whole parallel chord lies below the tangent: \(c < 0\). Now repeat Stage 2 with \(c < 0\): the inequality reverses, \(\frac{\ln p}{p} < \frac{\ln q}{q}\), and so \(q^p > p^q\).■

Looking back

• Reduce to one function. Comparing \(a^b\) with \(b^a\) became comparing \(h(a)\) with \(h(b)\), where \(h(x) = \frac{\ln x}{x}\).

• A sketch with reasons. Each “by means of a sketch” was backed up by a precise fact: concavity of \(\ln\) (chords below the curve, tangents above it) and the sign of an intercept.

• The special role of \(e\). The tangent from the origin touches at \(x = e\), and that is the dividing point for all of these comparisons.

What’s really going on

All solutions of \(x^y = y^x\)

Which pairs of distinct positive reals satisfy \(x^y = y^x\)? A neat trick finds them all. Put \(y = tx\) with \(t > 1\); then \(x^{tx} = (tx)^x\) gives \(x^t = tx\), so \(x^{t-1} = t\) and \[x = t^{1/(t-1)}, \qquad y = t^{t/(t-1)}.\] Writing \(t = 1 + \frac1n\) gives \(x = \left(1 + \frac1n\right)^n\) and \(y = \left(1 + \frac1n\right)^{n+1}\). With \(n = 1\) this is \((2, 4)\); with \(n = 2\) it is \(\left(\frac94, \frac{27}{8}\right)\), and indeed \(\left(\frac94\right)^{27/8} = \left(\frac{27}{8}\right)^{9/4}\). As \(n \to \infty\) both tend to \(e\), the point where the two branches meet. The problem was discussed in letters between Daniel Bernoulli and Christian Goldbach in 1728, and later by Euler.

The most economical base

The function \(x^{1/x} = e^{h(x)}\) is also maximised at \(x = e\). This answers a curious question: if the “cost” of a number system is the base times the number of digits needed, the most economical base is \(e\), and among whole numbers, \(3\) (narrowly beating \(2\)). A few experimental computers, such as the Soviet Setun of 1958, have used base \(3\).

The thinking skill

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Further reading

• Eli Maor, e: The Story of a Number (Princeton University Press). The history of \(e\), including the equation \(x^y = y^x\).

• Paul Nahin, An Imaginary Tale (Princeton University Press). Not only about \(\sqrt{-1}\): full of curious calculations of exactly this kind.

• Michael Spivak, Calculus (Cambridge University Press). Chapter 11 covers the use of derivatives to understand the shape of graphs with complete rigour.