Problem:
Define binary operations ⋄ and ♡ by
a⋄b=alog7(b) and a♡b=alog7(b)1
for all real numbers a and b for which these expressions are defined. The sequence (an) is defined recursively by a3=3♡2 and
an=(n♡(n−1))⋄an−1
for all integers n≥4. To the nearest integer, what is log7(a2019)?
Answer Choices:
A. 8
B. 9
C. 10
D. 11
E. 12
Solution:
By definition, the recursion becomes an=⎝⎜⎜⎛nlog7(n−1)1⎠⎟⎟⎞log7(an−1)=nlog7(n−1)log7(an−1). By the change of base formula, this reduces to an=nlogn−1(an−1). Thus, we have logn(an)=logn−1(an−1). Thus, for each positive integer m≥3, the value of logm(am) must be some constant value k.
We now compute k from a3. It is given that a3=3⊘2=3log7(2)1, so k=log3(a3)=log3⎝⎜⎜⎛3log7(2)1⎠⎟⎟⎞= log7(2)1=log2(7).
Now, we must have log2019(a2019)=k=log2(7). At this point, we simply switch some bases around. For those who are unfamiliar with logarithms, we can turn the logarithms into fractions which are less intimidating to work with.
log2019loga2019=log2log7⟹log7loga2019=log2log2019⟹log7(a2019)=log2(2019)
We conclude that log7(a2019)=log2(2019)≈11, or choice (D)11.
OR
Using the recursive definition, a4=(4⋄3)⋄(3⋄2) or a4=(4m)k where m=log7(3)1 and k=log7⎝⎜⎜⎛3log7(2)1⎠⎟⎟⎞.
Using logarithm rules, we can remove the exponent of the 3 so that k=log7(2)log7(3). Therefore, a4=4log7(2)1, which is 4◯2.
We claim that an=n⊘2 for all n≥3. We can prove this through induction.
Clearly, the base case where n=3 holds.
an=(n◯(n−1))⋄((n−1)⊙2)
This can be simplified as an=(nlogn−1(7))⋄((n−1)log2(7)).
Applying the diamond operation, we can simplify an=nh where h=logn−1(7)⋅log7(n−1)log2(7). By using logarithm rules to remove the exponent of log7(n−1) and after cancelling, h=log7(2)1.
Therefore, an=nlog7(2)1=n◯2 for all n≥3, completing the induction.
We have a2019=2019log2(7). Taking log2019 of both sides gives us log2019(a2019)=log2(7). Then, by changing to base 7 and after cancellation, we arrive at log7(a2019)=log2(2019). Because 211=2048 and 210=1024, our answer is (D)11 .
The problems on this page are the property of the MAA's American Mathematics Competitions