2016 AIME II Problems/Problem 3: Difference between revisions

Revision as of 21:30, 16 May 2016

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 9: / Line 9: @@
 Solution by Shaddoll
+== See also ==
+{{AIME box|year=2016|n=II|num-b=2|num-a=4}}

2016 AIME II (Problems • Answer Key • Resources)
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2016 AIME II Problems/Problem 3: Difference between revisions

Revision as of 21:30, 16 May 2016

Solution

See also