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Mock AIME II 2012 Problems/Problem 12

Revision as of 02:19, 5 April 2012 by Testingtesting (talk | contribs) (Created page with "==Problem== Let <math>\log_{a}b=5\log_{b}ac^4=3\log_{c}a^2b</math>. Assume the value of <math>\log_ab</math> has three real solutions <math>x,y,z</math>. If <math>\frac{1}{x}+\fr...")

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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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