How do you evaluate #log_3 1#?

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Answer 1 · 2016-08-15T20:25:53

I found that it is equal to zero:

You may use the definition of log:
#log_ba=x#
so that: #a=b^x#

we now need to find our #x# in:
#log_3(1)=x#

using our definition we see that the only possible value for #x# is zero because:
#1=3^0#

Answer 2 · 2016-08-15T21:24:26

#log_3 1 = 0#

Written in log form, the question being asked is,"

"1 is which power of 3? " #3^0 =1#

Or

"Using a base of 3, what index will give 1 as the answer?"
#3^0 =1#

Log form and index form are interchangeable.

#log_3 1 = x " " rArr 3^x = 1#

#x = 0#