assuming the q is correct
log((log alog b)(log blog a)) = log1 = 0
0/something nonzero = 0
[log {(log a)/(log b)} x (log b) / (log a)] / log {(log b)/(log a)}
assuming the q is correct
log((log alog b)(log blog a)) = log1 = 0
0/something nonzero = 0
the question is different.....
u dont have log b / log a inside the bracket in the numerator...
mentioned that in my second post....
its like...
log of log a upon log b
upon
log of log b upon log a
nd the whole multiplied by
log b upon log a
bhavnao ko samjho.... :p