Question #efe6c

1 Answer
Aug 17, 2017

#ln e^7=7#
#log_3 (1/3)=-1#
#log (3*1/3)=0#

Explanation:

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#