2007 AMC 12B Problem 17

For $b\gt0,$ we have $b\lt10^b,$ hence $\log_{10}b\lt b.$ If $b=1,$ then $a=0.$ If $b\gt1,$ then $0\lt\dfrac{\log_{10}b}{b^2}\lt1,$ so $a$ could not be an integer. Hence $0\lt b\lt1,$ so $\log_{10}b\lt0$ and $a=\dfrac{\log_{10}b}{b^2}\lt0.$ Thus $a\lt0\lt b\lt1\lt\dfrac1b,$ and the middle value of the sorted set is $b.$ Thus, the correct answer is D.