How do you evaluate #log_32 (1/2)#?

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Answer 1 · 2016-08-07T13:55:33

#- 1/5#

route 1 use the definition of log
let #log_(32) (1/2) = y#

#implies 32^y = 1/2#

and #32 = 2^5#

#implies (2^5)^y = 1/2#

#implies 2^(5y) = 1/2#

invert the LHS
#implies 1/(2^(-5y)) = 1/2^1#

so #-5y = 1, y = - 1/5#

route 2 flip the base

#log_(32) (1/2)#

#= (log_(x) (1/2))/(log_(x) (32))#
#= (-log_(x) (2))/(log_(x) (32))#

we can choose x to be what we want and here it seems that 2 must be a good candidate

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

#=- (log_(2) (2))/(log_(2) (2^5))#

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