How do you calculate #log_11 (1/sqrt 11)#?

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Answer 1 · 2016-09-08T14:56:34

#log_11 ( 1/sqrt11)=-1/2#

The question being asked in this log form is..

#log_11 ( 1/sqrt11)#

"What index/power of 11 will give #1/sqrt11#?"

The answer is not immediately obvious, so let's use some of the laws of indices to change #1/sqrt11# into a different form.

#1/sqrt11 = 1/11^(1/2) = 11^-(1/2)#

#log_11 11^(-1/2)# is now obvious by inspection,
because we can see that 11 has the index #-1/2#

#log_11 ( 1/sqrt11)= -1/2#

We could also get to the same result by using the change of base law.

#log_11 ( 1/sqrt11) = (log (1/sqrt11))/log11 = -1/2#