Difference between revisions of "2016 AIME II Problems/Problem 3"

Revision as of 16:16, 22 March 2018

Problem

Let $x,y,$ and $z$ be real numbers satisfying the system $\log_2(xyz-3+\log_5 x)=5$ , $\log_3(xyz-3+\log_5 y)=4$ , $\log_4(xyz-3+\log_5 z)=4$ , Find the value of $|\log_5 x|+|\log_5 y|+|\log_5 z|$ .

Solution

First, we get rid of logs by taking powers: $xyz-3+\log_5 x=2^{5}=32$ , $xyz-3+\log_5 y=3^{4}=81$ , and $(xyz-3+\log_5 z)=4^{4}=256$ . Adding all the equations up and using the $\log {xy}=\log {x}+\log{y}$ property, we have $3xyz+\log_5{xyz} = 378$ , so we have $xyz=125$ . Solving for $x,y,z$ by substituting $125$ for $xyz$ in each equation, we get $\log_5 x=-90, \log_5 y=-41, \log_5 z=134$ , so adding all the absolute values we have $90+41+134=\boxed{265}$ .

Solution by Shaddoll

@@ Line 1: / Line 1: @@
+==Problem==
 Let <math>x,y,</math> and <math>z</math> be real numbers satisfying the system
 <math>\log_2(xyz-3+\log_5 x)=5</math>,

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Difference between revisions of "2016 AIME II Problems/Problem 3"

Revision as of 16:16, 22 March 2018

Problem

Solution

See also