The best approach would be to find a value for each possible answer to compare with.
#"a)"# #0.02^0.03#
Let's try and do some calculations:
#0.02^0.03 = 0.02^(3/100)=root(100)(0.02^3)=root(100)(0.000008)#
What number do we have to raise to the #100#-th power to get a very small quantity? Well, it has to be smaller than #1#, and in fact, our answer is somewhat close to #1#.
To explain why, it'd be best to do an example. Take the number #0.90#, and let's raise it to some powers:
#0.90^2 =0.81#
#0.99^3 = 0.729#
#vdots#
#0.90^20 ~~0.12157#
As you can see, it gets smaller and smaller, yet it does so slowly.
#0.90^100 ~~ 0.00002#
While this is almost ten times bigger than #0.00008#, it can helps us realize something; if #n# is almost #1# then #n^k ~~0# for large #k#.
In other words, we have to raise a number close to #1# to the #100#-th power to get #0.000008#.
So #root(100)(0.000008) ~~ 1#, for all we care.
b) #0.03^0.02#.
This is the same story as a); it is very close to #1#.
c) #log_(0.98)1.01#
Let this quantity be #x#.
#log_(0.98) 1.01 =x=> 0.98^x =1.01#
You can observe from what we did before that if #x>1# then #0.98^x<0.98#, so to get a number bigger than 0.98, quite simply, #x# has to be negative. We can and will prove this below:
Say #x=-y#, for a positive number #y#.
#0.98^x=0.98^-y=1/0.98^y#
As #y# is positive, #0.98^y <1# and #1"/"0.98^y >1#.
Hence, #log_(0.98) 1.01<0#.
d) #sin(0.02)#
As #0.02# is pretty small, we could compare it to #0#:
#sin(0.02)~~sin0=0=>sin(0.02)~~0#
Also, #sin(0.02)>0#.
Conclusion
The numbers #0.02^0.03# and #0.03^0.02# are close to #1#, #log_(0.98)1.01# is negative and #sin(0.02)# is positive. Thus, the smallest number of the bunch is #log_(0.98)1.01#.
