2016 AMC 12A Problem 16

Let $u=\log_3 x.$ Then $\log_x 3=\dfrac1u,$ $\log_{\frac{1}{3}}x=-u,$ and $\log_x\dfrac13=-\dfrac1u.$ Two graphs meet where two of $u,\dfrac1u,-u,-\dfrac1u$ are equal for some valid $x\gt 0.$ Setting $u=\dfrac1u$ gives $u=\pm1,$ so $x=3$ or $x=\dfrac13;$ setting $-u=-\dfrac1u$ gives the same values. Setting $u=-u$ gives $u=0,$ i.e. $x=1,$ where $\log_3 x$ and $\log_{\frac{1}{3}}x$ are both $0.$ The remaining pairings have no real solution. The distinct intersection points are $(1,0),$ $(3,1),$ $\left(\dfrac13,-1\right),$ $(3,-1),$ and $\left(\dfrac13,1\right),$ so there are $5.$ Thus, the correct answer is D.