log5
(1/125)
Solution:
Since 5 is
under log , we pressume log 5 is as good as logarithm of 5 to the base
10.
Then log 5 (1/125) = log5+ log (1/125) = log5 + log
(1/5^3), log(ab) = log a+logb , log (a/b) = log a- logb , logarithm
properties.
= log5+log(1) - log5^3
=
= log5 +0 -3log5, as log(1) = 0 and log a^b =
bloga.
=log5 -3log 5
=
-2log5
= - 2*0.690970004
=
-1.397940009.
If the 5 is base of logarithm, then log5
(1/125) = log5 (1/5^3) = log5 (5^(-3) ) = -3 , as logk (k^m) = m, by definition of
logarithm.
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