How do you evaluate #log_[16]1 #?

2 Answers
Mar 22, 2016

Answer:

#log_16 1=0#

Explanation:

#log_16 1#
#=log 1/log 16#-> use change of base property #log_bx=logx/logb#
#=0/log 16#->#log 1=0#
#=0#

Mar 22, 2016

Answer:

0

Explanation:

By definition #log 1# to any base is equal to #0#.

Proof:
Let #log_16 1=n#
By definition of logarithmic function

#16^n=1#, This is true only if #n=0#