54 Let $a$, $b$, and $c$ sequentially represent the roots of the equations $x+\log _{2} x=2$, $x+\log _{3} x=2$, and $x+\log _{2} x=1$, respectively. Then their size relationship is
A. $a>b>c$
B. $b>a>c$
C. $c>a>b$
D. $c>b>a$
54 B. Obviously, $a, b, c \geqslant 1$. $a + \log_{3} a < a + \log_{2} a = 2 = b + \log_{3} b$, and $x + \log_{3} x$ is an increasing function of $x$, so $a < b$. Similarly, from $c + \log_{2} c = 1 < 2 = a + \log_{2} a$ we get $c < a$.