How do you evaluate #log_(1/2) (1/32)#?

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Answer 1 · 2016-08-12T12:47:31

1/5

Use #log_b a=log_c a/log_c b#

Here,

#log_(1/2) (1/32)#

#=log(1/32)/log(1/2)#

#=(-log 2)/(-log 32)#

#=log 2/log 2^5#

#=log 2/(5 log 2)#

#=1/5#