Problem:
Which of the following is the value of log2β6+log3β6β ?
Answer Choices:
A. 1
B. log5β6β
C. 2
D. log2β3β+log3β2β
E. log2β6β+log3β6β
Solution:
Recall that:
logbβ(uv)=logbβu+logbβvβ
logbβuβ
loguβb=1. We use these properties of logarithms to rewrite the original expression:
log2β6+log3β6β=(log2β2+log2β3)+(log3β2+log3β3)β
=2+log2β3+log3β2β
=(log2β3β+log3β2β)2β
=(D)log2β3β+log3β2ββ
OR
First,
log2β6+log3β6β=log2log6β+log3log6ββ=log3β
log2log6β
log3+log6β
log2ββ=log2β
log3log6(log2+log3)ββ
From here,
log2β
log3log6(log2+log3)ββ=log2β
log3(log2+log3)(log2+log3)ββ=log2β
log3(log2)2+2β
log2β
log3+(log3)2ββ
Finally,
log2β
log3(log2)2+2β
log2β
log3+(log3)2ββ=log2β
log3(log2+log3)2ββ
=log2β
log3βlog2β+log2β
log3βlog3β
=log3log2β+log2log3βββ
=log3β2β+log2β3β
Answer:
(D)log2β3β+log3β2ββ
The problems on this page are the property of the MAA's American Mathematics Competitions