log2k(N)|(1/a) = log2k(N)|a

log(N)|k¬(a) = klog(N)|a

log(N)|(a)k = 1/k.log(N)|a

log(N)*|(a)* = logN|a (¥ * khac 0)

logN|b = log(N|a)/log(b|a)

neu a>b>1 N>1 thi

logN|b>logN|a

giai bpt xet 2 TH a>1

0<a<1