The Beautiful and Useful Applications of Logarithms

Logarithms are among the most useful tools we have at our disposal in mathematics. They allows us to translate problems of a multiplicative nature to problems of an additive nature, and this can often be the key to unlocking a solution to a problem. They are especially useful for problems involving powers or indices.

To illustrate this, let’s look at an interesting number theory problem. We are looking to find all positive integer pairs n and m which satisfy the equation n^m = m^n, when n and m are distinct. If we play around with small n and m we can quickly see that 2⁴ = 16 = 4², so the pair 2 and 4 are certainly one solution.

It turns out that the pair 2 and 4 is in fact the only such solution. The more interesting aspect of this problem is how we prove that this is the only solution. The proof makes great use of logarithms and reduces to an analysis of a function which is continuous in the positive domain, rather than a discrete number theory approach. Let’s do it.

Turning the problem into a function to analyze

If we do some simple manipulation of our equation through taking natural logarithms, we can state it in an alternate form, as follows:

So a solution to our original equation represents two distinct positive integers n and m for which this function has the same value:

Sketching the function

Let’s sketch this function and see what it looks like. First, we can differentiate it and set the derivative to zero to determine if there are any stationary points on the curve.

This only solution to this is lnx = 1, so the function has a single stationary point at x = e, y = 1/e. Note also that

• At x = 1, y = 0

• As x approaches zero, lnx approaches -∞ and so y approaches -∞ also.

• As x gets large, and because lnx increases much more slowly than x, we have that y approaches zero.

Putting these facts together, we can conclude that our stationary point is a maximum, and we can sketch our function as follows, with the function crossing the x axis at x = 1:

So how is this useful?

Well , we can see from our sketch that the only area where our function can have the same value for two distinct values of x is where 0 < y < 1/e and x > 1.

We can also see that, if two such integer values of x do exist, then the smaller of the two must occur before our function hits its maximum. So let’s say n is the smaller of any integer pair that satisfies the equation. Then we can say that 1 < n < e.

Now, since e is somewhere between 2 and 3, we have to conclude that n must be 2. Further, from our curve, we can see that there can only be one other value of x for which (lnx)/x = (ln2)/2, and since we know that 2⁴= 4², we know that the other integer must be 4. Hence 2 and 4 must be the only pair of distinct integers which satisfy our equation.

If you have any interesting observations about the approach to this problem, feel free to mention them in the comments.