Problem:
The increasing geometric sequence x0​,x1​,x2​,… consists entirely of integral powers of 3. Given that
n=0∑7​log3​(xn​)=308 and 56≤log3​(n=0∑7​xn​)≤57
find log3​(x14​).
Solution:
The sequence is geometric, so there exist numbers a and r such that xn​=arn. It follows that
308=n=0∑7​log3​(xn​)=n=0∑7​log3​(arn)=n=0∑7​[log3​(a)+nlog3​(r)]=8log3​(a)+(n=0∑7​n)log3​(r)=8log3​(a)+28log3​(r).
Thus 2log3​(a)+7log3​(r)=77. Furthermore,
log3​(n=0∑7​xn​)=log3​(n=0∑7​arn)=log3​(a⋅r−1r8−1​)=log3​(ar7⋅1−r1​1−r81​​)=log3​(a)+7log3​(r)+log3​(1−r1​1−r81​​)​
which is between 56 and 57. Because the terms are all integral powers of 3, it follows that a and r must be powers of 3. Also, the sequence is increasing, so r is at least 3. Therefore
1=1−r1​1−r1​​<1−r1​1−r81​​<1−31​1​=23​<3, and 0<log3​(1−r1​1−r81​​)<1.
Also note that since a and r are powers of 3,log3​(a)+7log3​(r) is an integer and therefore must equal 56. Thus log3​(a)+7log3​(r)=56. The two equations log3​(a)+7log3​(r)=56 and 2log3​(a)+7log3​(r)=77 have the solution log3​(a)= 21 and log3​(r)=5. It follows that log3​(x14​)=log3​(ar14)=log3​(a)+14log3​(r)=21+14⋅5=91​.
The problems on this page are the property of the MAA's American Mathematics Competitions