How do you evaluate #\log _ { 1/ 2} ( 4) + \log _ { 1/ 2} ( 2) - \log _ { 1/ 2} ( 5)#?

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Answer 1 · 2017-06-05T08:06:45

#log_(1/2)4+log_(1/2)2-log_(1/2)5=-0.678#

#log_(1/2)4+log_(1/2)2-log_(1/2)5#

= #log_(1/2)((4xx2)/5)#

= #log_(1/2)(8/5)#

= #log_(1/2)1.6#

= #log1.6/log0.5#

= #0.20412/(-0.30103)#

= #-0.678#