# 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 $13+24=\boxed{037}$.

## Solution 2

Let $\log_a b = x$ and $\log_b c=y$, where $x,y>0$. Then, it is obvious that $log_c a = \frac{1}{xy}$.

We first focus on the first equality: $\log_a b = 5 \log_b{ac^4}$. This may be simplified using our logarithmic properties:

$$x = 5(\log_b a + \log_b c^4)$$

$$x = 5(\frac{1}{x} + 4y)$$

$$x = \frac{5}{x} + 20y.$$

Now, let's focus on the last expression: note that,

$$3 \log_c{a^2b} = 3 (2 \log_c a + \log_c b) = 6 \left(\frac{1}{xy}\right) + \frac{3}{y}.$$

We can equate all of these expressions:

$$x = \frac{5}{x} + 20y = 6 \left(\frac{1}{xy}\right) + \frac{3}{y}.$$

Multiplying all expressions by $xy$ gives us

$$x^2y = 5y + 20xy^2 = 6+3x.$$

Now, from our first equality we obtain

$$x^2 y = 5y + 20xy^2.$$

Since $y \ne 0$, we may safely divide by $y$:

$$x^2 = 5+20xy$$

$$y = \frac{x^2-5}{20x}.$$

From the first and last expressions we have:

$$x^2y = 6+3x$$

$$y= \frac{6+3x}{x^2}.$$

Equating our expressions for $y$ gives

$$\frac{x^2-5}{20x}=\frac{6+3x}{x^2}$$

$$x^4 - 5x^2 = 120x+60x^2.$$

Since $x \ne 0$, we may safely divide by $x$:

$$x^3 - 5x = 120 + 60x$$

$$x^3 - 65x-120=0.$$

By Vieta's formulas, we must have $r_1r_2+r_2r_3+r_1r_3 = -65$ and $r_1r_2r_3 = 120$. Dividing the former by the latter gives

$$\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3} = -\frac{65}{120} = -\frac{13}{24}$$

and hence $m+n = 13+24 = \boxed{037}$.

~FIREDRAGONMATH16