How do you calculate #log _(1/2) (9/4)#?

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Answer 1 · 2016-08-31T14:42:01

#log_(1/2)(9/4)=-1.17#

Let #log_(1/2)(9/4)=x#, then

#(1/2)^x=9/4# or

#1/2^x=9/4# or

#2^x=4/9# or

#x=log_2(4/9)#

= #log_2 4-log_2 9#

= #2-log9/log2#

= #2-0.9542/0.3010#

= #2-3.17=-1.17#