Mock AIME II 2012 Problems/Problem 12

Problem

Let $\log_{a}b=5\log_{b}ac^4=3\log_{c}a^2b$. Assume the value of $\log_ab$ has three real solutions $x,y,z$. If $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

Solution

Let $log_{a}b=15x$. Then $log_{b}ac^4=3x$ and $\log_{c}a^2b=5x$. From this, we have the system

\[a^{15x}=b\] \[b^{3x}=ac^4\] \[c^{5x}=a^2b\]

Substituting the first equation into the second, we obtain

\[a^{45x^2}=ac^4\rightarrow a^{\frac{45x^2-1}{4}}=c\]

Plugging this into the third equation yields $a^{225x^3-5x}=a^{60x+8}$.

Thus, $225x^3-65x-8=0$. Note that our three real roots multiply to $\frac{8}{225}$. However, since $\log_{a}b=15x$, we need to multiply by $15^3$, so our $xyz$ is \[\frac{8}{225}\cdot 15^3=8\cdot 15=120\]

We need $xy+xz+yz$. Using vieta’s and making sure we count for each factor of $15$ we divided off, we have $15^2\cdot\frac{-65}{225}$.

Our answer is $-\frac{65}{8\cdot 15}=-\frac{13}{24}$, thus $13+24=\boxed{037}$.

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