How do you find the exact value of #log_2 (-16)#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer A. S. Adikesavan 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, ..# Answer link Related questions What is a logarithm? What are common mistakes students make with logarithms? How can a logarithmic equation be solved by graphing? How can I calculate a logarithm without a calculator? How can logarithms be used to solve exponential equations? How do logarithmic functions work? What is the logarithm of a negative number? What is the logarithm of zero? How do I find the logarithm #log_(1/4) 1/64#? How do I find the logarithm #log_(2/3)(8/27)#? See all questions in Logarithm-- Inverse of an Exponential Function Impact of this question 3683 views around the world You can reuse this answer Creative Commons License