Difference between revisions of "2010 AIME II Problems/Problem 5"

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== Solution ==  
 
== Solution ==  
Using the properties of logarithms, <math>\log_{10}xyz = 81</math> by taking the log base 10 of both sides, and <math>(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}x)= 468</math> by using the fact that <math>\log_{10}ab = \log_{10}a + \log_{10}b</math>. Through further simplification, we find that <math>\log_{10}x+\log_{10}y+\log_{10}z = 81</math>. It can be seen that there is enough information to use the formula <math>\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc</math>, as we have both <math>\ a+b+c</math> and <math>\ 2ab+2ac+2bc</math>, and we want to find <math>\sqrt {a^2 + b^2 + c^2}</math>. After plugging in the values into the equation, we find that <math>\(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2</math> is equal to <math>\ 6561 - 936 = 5625</math>. However, we want to find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>, so we take the square root of <math>\ 5625</math>, or <math>\boxed{75}</math>.
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Using the properties of logarithms, <math>\log_{10}xyz = 81</math> by taking the log base 10 of both sides, and <math>(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}z)= 468</math> by using the fact that <math>\log_{10}ab = \log_{10}a + \log_{10}b</math>.  
 +
 
 +
Through further simplification, we find that <math>\log_{10}x+\log_{10}y+\log_{10}z = 81</math>. It can be seen that there is enough information to use the formula <math>\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc</math>, as we have both <math>\ a+b+c</math> and <math>\ 2ab+2ac+2bc</math>, and we want to find <math>\sqrt {a^2 + b^2 + c^2}</math>.  
 +
 
 +
After plugging in the values into the equation, we find that <math>\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2</math> is equal to <math>\ 6561 - 936 = 5625</math>.  
 +
 
 +
However, we want to find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>, so we take the square root of <math>\ 5625</math>, or <math>\boxed{075}</math>.
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 +
==Solution 2==
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<math>a = \log{x}</math>
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 +
<math>b = \log{y}</math>
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 +
<math>c = \log{z}</math>
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<math>xyz = 10^{81}</math>
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<math>\log{xyz} = 81</math>
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<math>\log{x} + \log{y} + \log{z} = 81</math>
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<math>a + b + c = 81</math>
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<math>\log{x}(\log{yz}) + \log{y}\log{z} = \log{x}(\log{y} + \log{z}) + \log{y}\log{z} = ab + ac + bc = 468</math>
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 +
<math>a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 6561</math>
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<math>a^2 + b^2 + c^2 = 5625 = 75^2</math>
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<math>\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}</math>
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==Video solution==
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https://www.youtube.com/watch?v=Ix6czB_A_Js&t
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2010|num-b=4|num-a=6|n=II}}
 
{{AIME box|year=2010|num-b=4|num-a=6|n=II}}
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{{MAA Notice}}

Revision as of 17:56, 21 July 2020

Problem

Positive numbers $x$, $y$, and $z$ satisfy $xyz = 10^{81}$ and $(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468$. Find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$.

Solution

Using the properties of logarithms, $\log_{10}xyz = 81$ by taking the log base 10 of both sides, and $(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}z)= 468$ by using the fact that $\log_{10}ab = \log_{10}a + \log_{10}b$.

Through further simplification, we find that $\log_{10}x+\log_{10}y+\log_{10}z = 81$. It can be seen that there is enough information to use the formula $\ (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$, as we have both $\ a+b+c$ and $\ 2ab+2ac+2bc$, and we want to find $\sqrt {a^2 + b^2 + c^2}$.

After plugging in the values into the equation, we find that $\ (\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2$ is equal to $\ 6561 - 936 = 5625$.

However, we want to find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$, so we take the square root of $\ 5625$, or $\boxed{075}$.

Solution 2

$a = \log{x}$

$b = \log{y}$

$c = \log{z}$

$xyz = 10^{81}$

$\log{xyz} = 81$

$\log{x} + \log{y} + \log{z} = 81$

$a + b + c = 81$

$\log{x}(\log{yz}) + \log{y}\log{z} = \log{x}(\log{y} + \log{z}) + \log{y}\log{z} = ab + ac + bc = 468$

$a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 6561$

$a^2 + b^2 + c^2 = 5625 = 75^2$

$\sqrt{\log{x^2} + \log{y^2} + \log{z^2}} = \sqrt{a^2 + b^2 + c^2} = \boxed{075}$

Video solution

https://www.youtube.com/watch?v=Ix6czB_A_Js&t

See also

2010 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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