A Common Tangent

For what value of $a$ ($a>1$) does $a^x$ have a common tangent with its inverse, $\log_a x$?

Let $c$ be the $x$-coordinate of the point of tangency, so that the $y$-coordinate is $a^c$ and $\log_a c$. Setting these $y$-coordinates equal doesn't seem to get us anywhere.

So let's move to the tangent line. The derivative of $a^x$ is $a^x\cdot \ln a$, while the derivative of $\log_a x$ is $1/(x\cdot \ln a)$. Plugging in our common point, $x=c$, and setting the two slopes equal, we get $a^c\cdot \ln a=1/(c\cdot \ln a)$. Again, not obviously useful.

What gets me through this problem is thinking about it graphically. Since $a^x$ and $\log_a x$ are inverse functions, each is the mirror image of the other across the line $y=x$. This means that their common tangent line will, in fact, be that same line, $y=x$.

Therefore, the slope of the common tangent line will be 1. Our two ways to represent the slope are $a^c\cdot \ln a$ and $1/(c\cdot \ln a)$. Setting the second equal to 1 and solving for $c$, we get $c=1/(\ln a)$, which we could also write as $(\ln e)/(\ln a)=\log_a e$. The first way we represented the point of tangency was $(c,a^c)$, but since it is on the line $y=x$, this is also $(c,c)$, so $c=a^c$.

We're almost there. In $c=a^c$, replace the left-hand side with $1/(\ln a)$, from our slope calculations. On the right-hand side, replace $c$ with $\log_a e$. Then we get $1/(\ln a)=a^{\log_a e}=e$, so $1/e=\ln a$, and we see that $a=e^{(1/e)}$.

I'm not sure why I first asked this question. I was teaching calc 1 at the time, and probably looking for interesting homework problems. I seem to recall chickening out in assigning this as a problem. Perhaps some other semester. Either way, I thought it was cute.

Update 20081229: Apparently (according to the penguin dictionary of curious and interesting numbers) this value, $e^{(1/e)}$ is also the value of $x$ for which $x^{(1/x)}$ is a maximum. This isn't entirely correct, as the maximum occurs at $x=e$, and therefore $e^{(1/e)}$ is the maximum value. Wikipedia has a page.