First we must always keep in mind the definition of a log which is represented by:
#log_a b=c# means that #a^c=b#
The definition of #ln# is #log# to the base #e#:
#log_e x=ln x#
Also some properties of logs that we must keep in mind are:
A) #log_x x=1#
B) #log_x (a/b)=log_x a - log_x b#
C) #log_x a^b=b*log_x a#
Now for our simplifications:
1) #ln e^7#
#7*ln elarr #pulling the exponent to the front using property C
#7*1# using the definition of #ln# and property A
Result = 7
2a) Since you didn't format it I'm first assuming you want the answer to #log_3 (1/3)#
#log_3 1 - log_3 3larr# using property B
#log_3 1 - 1larr# using property A
#0 -1larr# using the definition of #log# and the fact that #x^0=1#
Result = #-1#
2b) now assuming you want #log (3*1/3)#
#log (3/3)#
#log 1#
Result = #0 larr# using the definition of #log# and the fact that #x^0=1#
