How do you find the exact value of #log_2 (-16)#?

Question

3855 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-26T07:03:40

#=4+i((2n+1)pi)/ln2, n = 0, +-1, +-2, +-3, ..#, where #i =sqrt(-1)#

#When x <=0, log_b x# has values that are complex.

Here,

#log_2(-16)#

#=ln (-16)/ln 2#

#=ln ((+-i)^2 2^4)/ln 2#

#= (2 ln (+-i) + 4ln 2)/ln2#

#=4+2/ln2 lne^(i(2n+1)pi/2), n = 0, +-1, +-2, +-3, ..#

#=4+i((2n+1)pi)/ln2, n = 0, +-1, +-2, +-3, ..#