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

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

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