2016 AIME II Problem 3

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3.

Let x,x, y,y, and zz be real numbers satisfying the system log2(xyz3+log5x)=5\log_2(xyz - 3 + \log_5 x) = 5 log3(xyz3+log5y)=4\log_3(xyz - 3 + \log_5 y) = 4 log4(xyz3+log5z)=4.\log_4(xyz - 3 + \log_5 z) = 4. Find the value of log5x+log5y+log5z.|\log_5 x| + |\log_5 y| + |\log_5 z|.

Answer: 265
Concepts:logarithmsystem of equationssubstitution
Difficulty rating: 2300
Solution:

Exponentiating each equation gives xyz3+log5x=25xyz - 3 + \log_5 x = 2^5 and similarly for the others, so xyz+log5x=35,xyz + \log_5 x = 35, xyz+log5y=84,xyz + \log_5 y = 84, xyz+log5z=259.xyz + \log_5 z = 259. Write x=5a,x = 5^a, y=5b,y = 5^b, z=5c,z = 5^c, so that xyz=5a+b+c.xyz = 5^{a+b+c}.

Adding the three equations gives 35s+s=378,3 \cdot 5^s + s = 378, where s=a+b+c.s = a + b + c. The left side is strictly increasing in s,s, and s=3s = 3 works since 375+3=378,375 + 3 = 378, so s=3s = 3 and 5s=125.5^s = 125.

Then a=35125=90,a = 35 - 125 = -90, b=84125=41,b = 84 - 125 = -41, and c=259125=134,c = 259 - 125 = 134, so a+b+c|a| + |b| + |c| =90+41+134= 90 + 41 + 134 =265.= 265.

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