Logarithm-exponential Equation Conundrum

The graphs in the problem are misleading. While they correctly show that both functions are decreasing and concave up, and symmetric about the $y=x$ line, the conclusion of having only one intersection is unfounded. Nothing prevents two decreasing, concave up graphs from intersecting in more than one point:

In fact, we will prove that there are three points of intersection: $x=\frac{1}{4}$ , $x=\frac{1}{2}$ and such $x$ that $\log_{1/16}x=x.$ To do this, consider the difference function: $f(x)=\log_{1/16}x-(1/16)^x$ . Its derivative, $f'(x)=-\frac{1}{x\ln 16}+(1/16)^x\ln 16$ . So $f'(x)=0$ if and only if $\left(\frac{1}{16} \right)^x = \frac{1}{x (\ln 16)^2}$ Taking logarithms of both sides, we get $-x\ln 16=-\ln x -2\ln (\ln 16)$ This can be rewritten as $\ln x = (\ln 16)x + \ln (\ln 16)$ . The left hand side is concave down, while the right hand side is linear, so they have no more than two points of intersection. Therefore, $f(x)$ can have no more than three zeroes. We already know two of them: $x=\frac{1}{4}$ and $x=\frac{1}{2}$ , and from the Intermediate Value Theorem applied to the function $\log_{1/16}x-x,$ there exists $c$ such that $\log_{1/16}c=c.$ For this $c$ we have $(1/16)^c=c,$ so $f(c)=0.$

So, the original equation has exactly $3$ solutions.

The word "sketch" makes me suspicious of the use of the graph... The algebra does not lie!

Moreover, the solutions given in the problem are mirror images of each other: $(\tfrac12, \tfrac14)$ and $(\tfrac14, \tfrac12)$ .

But if the graphs pass through the line of symmetry (and there is no reason why they shouldn't), there must also be a solution that lies on that line, so that $f(x) = f^{-1}(x) = x$ . Therefore there are at least three solutions.

I was too lazy to prove that there were only three solutions. Calvin gives the argument nicely in terms of convexity.