比較兩個數的大小

Let f(x)= ln(x) /x, then f'(x)= [1- ln(x) ]/ x²
0 < x < e, f'(x) > 0, f(x) is increasing.
e < x, f'(x) < 0, f(x) is decreasing.
thus f(2005) > f(2006)
ln(2005) / 2005 > ln(2006) / 2006
2006 ln(2005) > 2005 ln(2006)
ln( 2005^2006) > ln( 2006^2005)
so that 2005^2006 > 2006^2005

2005 > e > (1 + 1/2005)^2005
=> 2005^2006 > 2006^2005

Roughly speaking, if you do MC question in exam similar to this question,
you can directly take log for both expressions, giving:
log 2005^2006 = 2006 log 2005 = 15252.41...
log 2006^2005 = 2005 log 2006 = 15245.81...
Since
log 2005^2006 > log 2006^2005
2005^2006 > 2006^2005