AMC12 2013 B

Answer (A): Using the change of base identity gives $\log n\cdot \log_n x=\log x$ and $\log m\cdot \log_m x=\log x$. The equivalent equation is $$(\log x)^2-\frac18(7\log m+6\log n)\log x-\frac{2013}{8}\log m\cdot \log n=0.$$ As a quadratic equation in $\log x$, the sum of the two solutions $\log x_1$ and $\log x_2$ is equal to the negative of the linear coefficient. It follows that $$\log(x_1x_2)=\log x_1+\log x_2=\frac18(7\log m+6\log n)=\log\left((m^7n^6)^{1/8}\right).$$ Let $N=x_1x_2$ be the product of the solutions. Suppose $p$ is a prime dividing $m$. Let $p^a$ and $p^b$ be the largest powers of $p$ that divide $m$ and $n$ respectively. Then $p^{7a+6b}$ is the largest power of $p$ that divides $m^7n^6=N^8$. It follows that $7a+6b\equiv 0\pmod 8$. If $a$ is odd, then there is no solution to $7a+6b\equiv 0\pmod 8$ because $7a$ is not divisible by $\gcd(6,8)=2$. If $a\equiv 0\pmod 8$, then because $a>0$, it follows that $N^8=m^7n^6\ge (p^8)^7=p^{56}\ge 2^{56}$, so $N\ge 2^7=128$. If $a\equiv 2\pmod 8$ then $14+6b\equiv 0\pmod 8$ is equivalent to $3b+3\equiv 3b+7\equiv 0\pmod 4$. Thus $b\equiv 3\pmod 4$ and then $N^8=m^7n^6\ge (p^2)^7(p^3)^6=p^{32}\ge 2^{32}$, so $N\ge 2^4=16$ with equality for $m=2^2$ and $n=2^3$. Finally, if $a\ge 4$ and $a$ is not a multiple of $8$, then $b\ge 1$ and thus $N^8=m^7n^6\ge (p^4)^7(p^1)^6=p^{34}\ge 2^{34}$, so $N\ge 2^{17/4}>2^4=16$. Therefore the minimum product is $N=16$ obtained uniquely when $m=2^2$ and $n=2^3$. The requested sum is $m+n=4+8=12$.