Problem:
Let m>1 and n>1 be integers. Suppose that the product of the solutions for x of the equation
8(lognx)(logmx)−7lognx−6logmx−2013=0
is the smallest possible integer. What is m+n ?
Answer Choices:
A. 12
B. 20
C. 24
D. 48
E. 272
Solution:
Using the change of base identity gives logn⋅lognx=logx and logm⋅logmx= logx. The equivalent equation is
(logx)2−81(7logm+6logn)logx−82013logm⋅logn=0
As a quadratic equation in logx, the sum of the two solutions logx1 and logx2 is equal to the negative of the linear coefficient. It follows that
log(x1x2)=logx1+logx2=81(7logm+6logn)=log((m7n6)1/8).
Let N=x1x2 be the product of the solutions. Suppose p is a prime dividing m. Let pa and pb be the largest powers of p that divide m and n respectively. Then p7a+6b is the largest power of p that divides m7n6=N8. It follows that 7a+6b≡0(mod8). If a is odd, then there is no solution to 7a+6b≡0(mod8) because 7a is not divisible by gcd(6,8)=2. If a≡0(mod8), then because a>0, it follows that N8=m7n6≥(p8)7=p56≥256, so N≥27=128. If a≡2(mod8) then 14+6b≡0(mod8) is equivalent to 3b+3≡3b+7≡0 (mod4). Thus b≡3(mod4) and then N8=m7n6≥(p2)7(p3)6=p32≥232, so N≥24=16 with equality for m=22 and n=23. Finally, if a≥4 and a is not a multiple of 8 , then b≥1 and thus N8=m7n6≥(p4)7(p1)6=p34≥234, so N≥217/4>24=16. Therefore the minimum product is N=16 obtained uniquely when m=22 and n=23. The requested sum is m+n=4+8=12.
The problems on this page are the property of the MAA's American Mathematics Competitions