How do you evaluate #log_14 (-1)#?

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Answer 1 · 2016-08-09T16:26:52

#((4n+1)pi)/ln 14 i, n = 0, +-1, +-2. +-3, ...#

The values are complex.

#log_14 (-1)#

#=ln(-1)/ln 14#

#=ln(i^2)/ln 14#

#=2 ln i /ln 14#

#=2lne^(i(2npi+pi/2))/ln 14, n = 0, +-1. +-2, +-3, ..#.

#=((4n+1)pi i) /ln 14, n = 0, +-1. +-2, +-3, ..#

Answer 2 · 2016-08-09T16:31:09

#log_{14}(-1) = i (pi pm 2kpi )/log_e 14#

for #k = 0,1,2, cdots#

Let us investigate the complex solutions. We know by the logarithm definition

#14^z = -1#

Supposing now #z = x + i y# we have

#14^x 14^{iy} = -1# so we have

#14^x = 1-> x = 0# and
#14^{iy} = -1#

We know

#14 = e^{log_e 14}# so

#14^{iy} = e^{i y log_e 14} = cos(y log_e14)+i sin(y log_e 14) = -1#.

(We used de Moivre's identity #e^{i phi)=cos(phi)+i sin(phi)#)

This condition is attained for

#y log_e14 = pi pm 2kpi# with # k= 0,1,2,3,4,cdots#

so

#y = (pi pm 2kpi )/log_e 14#

Finally

#log_{14}(-1) = i (pi pm 2kpi )/log_e 14#