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

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Answer 1 · 2016-03-22T16:45:22

#log_16 1=0#

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

Answer 2 · 2016-03-22T17:22:27

0

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#