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

1 Answer
Jan 26, 2017

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

Explanation:

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, ..