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

Revision as of 16:56, 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

@@ Line 36: / Line 36: @@
 ==Video solution==
-https://www.youtube.com/watch?v=Ix6czB_A_Js&t=2s
+https://www.youtube.com/watch?v=Ix6czB_A_Js&t
 == See also ==
 {{AIME box|year=2010|num-b=4|num-a=6|n=II}}
 {{MAA Notice}}

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