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

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== See also ==
== See also ==
{{AIME box|year=2016|n=II|num-b=2|num-a=4}}
{{AIME box|year=2016|n=II|num-b=2|num-a=4}}
{{MAA Notice}}

Revision as of 20:15, 18 August 2020


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 $|\log_5 x|+|\log_5 y|+|\log_5 z|$.


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

See also

2016 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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