This problem was part of a Montgomery County Math competition a few years ago, and I’ve had it posted on my whiteboard for the past two weeks:

Solve for .

If you haven’t at least tried it, go ahead and do it now. Then check out the solutions below. Sorry it’s so long!

**Here’s one solution** (stay with me on this!). Rewrite in log form and do some algebra, applying some log properties:

Notice the above equation is quadratic in . By the quadratic formula, we have our solutions (albeit, unrecognizable right now),

Now, this gets a bit tricky. Watch the magic under the radical unfold.

Did you see that? That was amazing! We wrote the discrimant as a perfect square. Pretty cool. Let’s not get too excited though. Finishing the problem:

or

or

or

or

or

or

or

So that’s how I got the two answers.

**Arun’s Solution.** Now, I’d like to share how one of my students, Arun, solved the problem in a slightly more elegant (or quicker, at least) way. First, we introduce a new variable,

and

and

Let’s work with the first equation:

Now substitute the second equation :

So we see that either

or

or

or

or

Cool, right?? I love math.