2016 AIME II Problems/Problem 3

Revision as of 12:40, 15 March 2021 by Jske25 (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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 $90+41+134=\boxed{265}$.

Note: $xyz=125$ because we know $xyz$ has to be a power of $5$, and so it is not hard to test values in the equation $3xyz+\log_5{xyz} = 378$ in order to achieve desired value for $xyz$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS