∵log₃1＜a=log₃2＜log₃3∴0＜a＜1

∵b=log₂3＞log₂2=1∴b＞1

∵c=log

1

2

1

5

＞log

1

2

1

2

= 1∴c＞1

∵∵c=log

1

2

1

5

= log25＞log23=b      即c＞b

故有c＞b＞a

故选A