How To Really Solve This Tricky Algebra Problem? (III)

Welcome to the third entry in the tricky algebra problem series. When you take a look at this equation, it does indeed look tricky. Unlike the previous couple of equations, I must confess that this one is indeed tricky!

Furthermore, I am being a little bit naughty by classifying this as an algebra problem; you will need a bit more than just algebra to solve this one. But hey, I am at least being upfront about it!

Spoiler Alert:

If you are adventurous enough to solve this equation on your own, I welcome you to do so now. You may choose to come back and continue reading this essay once you are done. Beyond this section, I will be explicitly discussing solutions to the equation.

This essay is supported by Generatebg

A Crude Approach to Solving the Tricky Algebra Problem

As I have previously mentioned, my instinct to solving equations with unknown exponents is to dethrone the exponent(s) first. The logarithmic function is usually a good candidate for this purpose. Let us find out what happens if we choose to take this approach.

As you can see, we have x/log(x) on the left-hand side and y/log(y) on the right-hand side. This is not looking good. But still, the situation is not as hopeless as it appears to be. We could consider different values of ‘x’ and ‘y’ and choose to graph the respective output functions in the cartesian plane.

The intersecting point(s) of the two graphs would be solutions to the equation. However, there are several problems with this approach:

1. There are infinite pairs of ‘x’ and ‘y’ for us to consider.

2. Graphing is resource intensive.

3. We don’t need the logarithmic function to graph these quantities; we might have as well done it with the original equation.

Considering these disadvantages, it makes sense to abandon this approach and consider a more refined one to arrive at a generalised solution. And that is exactly what I’m choosing to do!

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A Refined Approach to Solving the Tricky Algebra Problem

A realization that would really help us is the fact that x^y is nothing but ‘x’ multiplied with itself ‘x’ number of times and y^x is nothing but ‘y’ multiplied with itself ‘x’ number of times.

What this means is that ‘x’ is a multiple of ‘y’ and ‘y’ is a multiple of ‘x’. In order to solve the equation, we could consider one of these facts and try to create a new equation. For instance, let us express ‘y’ as a multiple of ‘x’.

Here, the factor ‘t’ is called a parameter and can take any real number value. Furthermore, it is important to note that ‘t’ is independent of ‘x’ or ‘y’. By expressing ‘y’ as a multiple of ‘x’ using a parameter ‘t’, we have created a new equation.

In mathematics, such an equation is known as a parametric equation (which is quite commonly used in applied mathematics, engineering, etc.).

Parametric equations are equations that describe unknowns as explicit functions of independent ‘parameters’. For example, the following parametric equation set can be used to describe a circle:

x = r cos(t)

y = r sin(t)

With the new parametric equation, we have finally arrived at a stage where we can treat the problem algebraically.

How to Really Solve This Tricky Algebra Problem?

I am now going to plug in the parametric value of y (= tx) into the original equation and take the x-th root on both the sides of the resulting equation:

I am going to proceed with algebraic manipulation with the aim of expressing ‘x’ exclusively in terms of ‘t’ (or the other way around).

We could now choose to multiply by t^(1/t) and divide by x^(1/t) on both sides, to proceed as follows:

There we go. We have arrived at an equation that expresses ‘x’ exclusively in terms of ‘t’ (my original goal).

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The Solution to the Tricky Algebra Problem

We are almost there. All that is left is to plug in this equation into the original parametric equation and express ‘y’ exclusively in terms of ‘t’ as well.

We can now express the solution to our original equation as a set of parametric equations as follows:

There are a few last points that we need to address. When t=1, we end up in a situation where the exponent for both x and y is 1/0. In mathematics, division by zero is undefined (for details, refer to my essay on what really happens when you divide by zero).

Similarly, when t = 0, we end up with 0⁰, which is also undefined for a different set of reasons as compared to division by zero (for details, refer to my essay on what really happens when you raise zero to the power of zero).

Finally, as reader Jasmin Tremblay pointed out, in order to restrict the solution domain to the real number line (and avoid the complex plane), we need to restrict t values to the domain: t>1 (t=0 and t=1 lead to undefined states as discussed above).

Sample Solution for the Tricky Algebra Problem

Just to test out the results, let us consider ‘t = 4’:

As you can see, our solution set of parametric equations works for any real value of t, which includes rational and irrational numbers.

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Further reading that might interest you: How To Really Solve This Tricky Algebra Problem? and How To Really Solve i^i?

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