# A Good Log Problem

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 $x$. $\displaystyle x^{\log 5x}=2$

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: $\displaystyle \log 5x=\log_x 2$ $\displaystyle \log 5 + \log x=\frac{\log 2}{\log x}$ $\displaystyle (\log 5)(\log x) + (\log x)^2= \log 2$ $\displaystyle (\log x)^2 + (\log 5)(\log x) - \log 2 = 0$

Notice the above equation is quadratic in $\log x$. By the quadratic formula, we have our solutions (albeit, unrecognizable right now), $\displaystyle \log x = \frac {-\log 5 \pm \sqrt{(\log 5)^2-4(1)(-\log 2)}}{2(1)}$ $\displaystyle = \frac {-\log 5 \pm \sqrt{(\log 5)^2+4(\log 2)}}{2}$

Now, this gets a bit tricky. Watch the magic under the radical unfold. $\displaystyle =\frac {-\log 5 \pm \sqrt{(\log 5)^2+4(\log (10/5))}}{2}$ $\displaystyle =\frac {-\log 5 \pm \sqrt{(\log 5)^2+4(\log 10 - \log 5))}}{2}$ $\displaystyle =\frac{-\log 5 \pm \sqrt{(\log 5)^2+4(1 -\log 5)}}{2}$ $\displaystyle =\frac {-\log 5 \pm \sqrt{(\log 5)^2+4 -4(\log 5)}}{2}$ $\displaystyle =\frac {-\log 5 \pm \sqrt{(\log 5)^2-4(\log 5)+4}}{2}$ $\displaystyle =\frac {-\log 5 \pm \sqrt{(\log 5 - 2)^2}}{2}$ $\displaystyle =\frac {-\log 5 \pm (\log 5 - 2)}{2}$

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: $\displaystyle \log x = \frac {-\log 5+\log 5 - 2}{2}$ or $\displaystyle \log x = \frac {-\log 5-\log 5+2}{2}$ $\displaystyle \log x = \frac {-2}{2}$ or $\displaystyle \log x = \frac {-2\log 5+2}{2}$ $\displaystyle \log x =-1$ or $\displaystyle \log x =-\log 5+1$ $\displaystyle \log x =-1$ or $\displaystyle \log x =-\log 5+\log 10$ $\displaystyle \log x =-1$ or $\displaystyle \log x=-\log (10/5)$ $\displaystyle \log x =-1$ or $\displaystyle \log x =-\log 2$ $\displaystyle x=\frac{1}{10}$ or $\displaystyle x=2$

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, $a$ $\displaystyle \log 5x=\log_x 2=a$ $\displaystyle \log 5x=a$   and $\displaystyle \log_x 2=a$ $\displaystyle 10^a=5x$  and $\displaystyle x^a=2$

Let’s work with the first equation: $\displaystyle 10^a=5x$ $\displaystyle 10\cdot10^{a-1}=5x$ $\displaystyle 2\cdot10^{a-1}=x$

Now substitute the second equation $x^a=2$: $\displaystyle x^a\cdot10^{a-1}=x$ $\displaystyle x^{a-1}\cdot10^{a-1}=1$ $\displaystyle (10x)^{a-1}=1$

So we see that either $\displaystyle 10x=1$  or $\displaystyle a-1=0$ $\displaystyle x=\frac{1}{10}$  or $\displaystyle a=1$ $\displaystyle x=\frac{1}{10}$  or $\displaystyle \log_x 2=1$ $\displaystyle x=\frac{1}{10}$  or $\displaystyle x=2$

Cool, right?? I love math.