Problem:
Two non-zero real numbers, a and b, satisfy ab=aβb. Find a possible value of baβ+abββab.
Answer Choices:
A. β2
B. β21β
C. 31β
D. 21β
E. 2
Solution:
Find the common denominator and replace the ab in the numerator with aβb to get
baβ+abββabβ=aba2+b2β(ab)2β=aba2+b2β(aβb)2β=aba2+b2β(a2β2ab+b2)β=ab2abβ=2.β
OR
Note that a=a/bβ1 and b=1βb/a. It follows that baβ+abββab=(a+1)+ (1βb)β(aβb)=2.
Answer: Eβ.
The problems on this page are the property of the MAA's American Mathematics Competitions