How do you evaluate #log_7 (-1)#?

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Answer 1 · 2016-09-27T07:30:12

Principal value is #(pi/ln 7)i#
The general value is
#2(2kpi+pi/2)i/ln 7, k = 0, +-1, +-2, +-3, ...#

Use #i&2=-1, i=e^(pi/2i) and log _b a = ln a/ln b and ln a^2=2 ln a #.

Now,

#log_7(-1)#

#=ln(-1)/ln 7#

#=ln i^2/ln 7#

#=2 ln i/ln 7#

#=2ln e^(pi/2i)/ln 7#

#=(2)(pi/2)i/ln 7#

#=(pi/ln 7)i#

This is the principal value.

The general value is

#2(2kpi+pi/2)i/ln 7, k = 0, +-1, +-2, +-3, ...#