2013 AMC 12B Problem 22

Writing $\log_n x = \tfrac{\log x}{\log n}$ and $\log_m x = \tfrac{\log x}{\log m},$ the equation becomes a quadratic in $\log x$ whose roots sum to $\log(x_1 x_2)$ $= \tfrac18(7\log m + 6\log n).$ Hence $N^8=m^7n^6,$ where $N=x_1x_2.$ For each prime dividing $mn,$ let its exponents in $m,n$ be $a,b.$ Then $7a+6b\equiv0\pmod8.$ An odd $a$ is impossible. If $a=0,$ then $b$ is a multiple of $4$ and this prime contributes at least $p^3,$ but some other prime must divide $m.$ If $a\equiv0\pmod8$ is positive, its contribution to $N$ is at least $p^7;$ if $a\equiv2\pmod8,$ then $b\equiv3\pmod4$ and the least contribution is $p^4.$ Every other positive even $a$ gives more. Thus the minimum uses only $p=2$ with $(a,b)=(2,3),$ giving $N=16,$ $m=4,$ and $n=8.$ So $m+n=12.$ Thus, the correct answer is A.