2010 AIME II Problems/Problem 5

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 $(\log_{10}x)(\log_{10}y) + (\log_{10}x)(\log_{10}z) + (\log_{10}y)(\log_{10}x)= 468$ 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 $\boxed{075}$ .

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2010 AIME II Problems/Problem 5

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