AMC 12 Mock Exam 3 Problem 19

Let $x=\log a$, $y=\log b$, $z=\log c$ (all positive) and $S=x+y+z$. Then $\log_a(bc)=\frac{y+z}{x}=\frac{S-x}{x}=2$ gives $S=3x$, and $\log_b(ca)=\frac{S-y}{y}=3$ gives $S=4y$. Hence $x=\frac S3$, $y=\frac S4$, $z=S-\frac S3-\frac S4=\frac{5S}{12}$, and $$\log_c(ab)=\frac{x+y}{z}=\frac{\frac{7S}{12}}{\frac{5S}{12}}=\boxed{\tfrac75}.$$ (Such numbers exist: for example $a=e^4$, $b=e^3$, $c=e^5$. In general $\frac1{1+\log_a(bc)}+\frac1{1+\log_b(ca)}+\frac1{1+\log_c(ab)}=\frac xS+\frac yS+\frac zS=1$.)