How do you evaluate #log_2 (1/16)#?

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Answer 1 · 2016-06-28T08:01:19

-4

let #z = log_2 (1/16)#

#\implies z = - log_2 (16)#

#\implies -z = log_2 (16)#

# = log_2 (2^4)#

# = 4 log_2 (2)#

# = 4*1 = 4#

#z = -4#