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Difference between revisions of "Mock AIME II 2012 Problems/Problem 12"

(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...")
 
m (→‎Solution 2)
 
(2 intermediate revisions by the same user not shown)
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Our answer is <math>-\frac{65}{8\cdot 15}=-\frac{13}{24}</math>, thus <math>13+24=\boxed{037}</math>.
 
Our answer is <math>-\frac{65}{8\cdot 15}=-\frac{13}{24}</math>, thus <math>13+24=\boxed{037}</math>.
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 +
==Solution 2==
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Let <math>\log_a b = x</math> and <math>\log_b c=y</math>, where <math>x,y>0</math>. Then, it is obvious that <math>log_c a = \frac{1}{xy}</math>.
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We first focus on the first equality: <math>\log_a b = 5 \log_b{ac^4}</math>. This may be simplified using our logarithmic properties:
 +
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<cmath>x = 5(\log_b a + \log_b c^4)</cmath>
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<cmath>x = 5(\frac{1}{x} + 4y)</cmath>
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<cmath>x = \frac{5}{x} + 20y.</cmath>
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Now, let's focus on the last expression: note that,
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<cmath>3 \log_c{a^2b} = 3 (2 \log_c a + \log_c b) = 6 \left(\frac{1}{xy}\right) + \frac{3}{y}.</cmath>
 +
 +
We can equate all of these expressions:
 +
 +
<cmath>x = \frac{5}{x} + 20y = 6 \left(\frac{1}{xy}\right) + \frac{3}{y}.</cmath>
 +
 +
Multiplying all expressions by <math>xy</math> gives us
 +
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<cmath>x^2y = 5y + 20xy^2 = 6+3x.</cmath>
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Now, from our first equality we obtain
 +
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<cmath>x^2 y = 5y + 20xy^2.</cmath>
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Since <math>y \ne 0</math>, we may safely divide by <math>y</math>:
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<cmath>x^2 = 5+20xy</cmath>
 +
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<cmath>y = \frac{x^2-5}{20x}.</cmath>
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From the first and last expressions we have:
 +
 +
<cmath>x^2y = 6+3x</cmath>
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<cmath>y= \frac{6+3x}{x^2}.</cmath>
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Equating our expressions for <math>y</math> gives
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<cmath> \frac{x^2-5}{20x}=\frac{6+3x}{x^2}</cmath>
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<cmath>x^4 - 5x^2 = 120x+60x^2.</cmath>
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Since <math>x \ne 0</math>, we may safely divide by <math>x</math>:
 +
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<cmath>x^3 - 5x = 120 + 60x</cmath>
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<cmath>x^3 - 65x-120=0.</cmath>
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By Vieta's formulas, we must have <math>r_1r_2+r_2r_3+r_1r_3 = -65</math> and <math>r_1r_2r_3 = 120</math>. Dividing the former by the latter gives
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<cmath>\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3} = -\frac{65}{120} = -\frac{13}{24}</cmath>
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and hence <math>m+n = 13+24 = \boxed{037}</math>.
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~FIREDRAGONMATH16

Latest revision as of 11:09, 8 July 2022

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

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