Here's an interesting mental math challenge. Given two different positive real numbers, which we'll call *a* and *b*, which is greater, *a*^{b} or *b*^{a}? If you're able to calculate both exponents mentally, such as 2^{3} vs 3^{2}, then that's probably the simplest way to go. What happens if one or both exponential expressions are too hard to mentally calculate?

To solve this, we'll need to find a general rule. Some advanced calculation will be required to find it. However, once we do the work to find the rule, you'll see that almost no math will be needed to solve these types of problems!

**SEARCH FOR THE RULE:** Consider that 2^{3} < 3^{2}, but 3^{4} > 4^{3}. Just looking at the placement of smaller and bigger numbers, there doesn't seem to hard and fast rule which applies. The best place to start is just by assuming what we wish to discover:

From that point, let's see if we can separate *a* and *b* somehow. The quickest way is to raise both sides to the power of 1/*ab*:

In English, then, when the *a*th root of *a* is greater than the *b*th root of *b*, then *a*^{b} > *b*^{a}. Don't worry, though. You won't have to do any roots in your head. Instead, just look at this Desmos graph where y=*x*th root of *x*. If you click in the box marked 1 at the upper left, Desmos highlights two points, the point with the minimum y value, which is (0, 0), and the point with the maximum value, (2.718, 1.445).

Wait a minute! The value of 2.718 sounds familiar. Could the maximum value of *x*th root of *x* really be *e*? Sure enough, Wolfram|Alpha verifies that the maximum is *e*!

Look at both sides of the graph, then. From 0 to *e*, the graph increases. From *e* on upwards, the value of the *x*th root of *x* will steadily decrease. In fact, the value will just keep getting closer and closer to 1.

This means we've found the start of our rule. When you have two numbers, both of which are equal to or greater than *e*, the smaller number *x* will always yield a greater *x*th root of *x* than the higher number. This tells us that the *a*th root of *a* is greater than the *b*th root of *b* when a is the smaller number. Working backwards to our original assumption, this means that:

There's our rule!Whenaandbare both equal to or greater thane,a^{b}is always greater thanb^{a}whenais the smaller number andbis the larger number.

**WORKING WITH INTEGERS:**What about when one or both of

*a*and

*b*are less than

*e*? In the case of positive integers, this means we only have to consider the cases involving 1 and 2.

The case of 1 is easy. Assume that

*a*equals 1 and

*b*has a value of 2 or more. 1

^{b}will always be 1, and

*b*

^{1}will always be

*b*, which is 2 or more. That covers every case involving 1.

What about 2? Let's work through each case individually, always assigning

*a*to a value of 2, and

*b*a value other than 2. If

*b*is 1, we just covered that case.

*b*can't be 2, as it would be the same as

*a*. What about when

*b*is 3? We already covered the fact that 2

^{3}< 3

^{2}, so we remember that special case. What happens when

*b*is 4? This is a very unique case, as 2

^{4}is exactly equal to 4

^{2}! In fact, this is a well-known special case, as it's the only time when

*a*

^{b}equals

*b*

^{a}when

*a*and

*b*are different integers. When

*b*is 5 ore greater, as you can see in this graph, we can fall back on the rule we set above.

**A CLASSIC CHALLENGE:**Which is greater, π

^{e}or

*e*

^{π}? There's a problem you probably never thought you'd solve in your head!

Let's go see if our primary rule applies. Are

*a*and

*b*both equal to or greater than

*e*?

*e*, of course, is exactly equal to

*e*(2.71828 and so on). π, as any Grey Matters readers should already know, is roughly 3.1415, and therefore also greater than

*e*.

This means we can apply our primary rule!

*e*is smaller than π, which means that

*e*

^{π}must be greater than π

^{e}. A quick verification with Wolfram|Alpha shows that this is correct.

About 4 years ago, Presh Talwalkar wrote up several approaches to this classic problem. Interestingly, this approach wasn't included. The 4th method does work with the

*x*th root of

*x*as we did, but it uses a deeper approach involving calculus.

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