Problem:
Suppose that a,b, and c are positive real numbers such that alog3β7=27, blog7β11=49, and clog11β25=11β. Find
a(log3β7)2+b(log7β11)2+c(log11β25)2
Solution:
It follows from the properties of exponents that
βa(log3β7)2+b(log7β11)2+c(log11β25)2=(alog3β7)log3β7+(blog7β11)log7β11+(clog11β25)log11β25=27log3β7+49log7β11+11βlog11β25=33log3β7+72log7β11+1121ββ
log11β25=73+112+25β=343+121+5=469β.β
The problems on this page are the property of the MAA's American Mathematics Competitions