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

1 Answer

Sep 25, 2016

#-(ln9-ln4)/ln2~=-1.16992#

Explanation:

Remember: #log_a x = log_b x/log_b a#

Let #a=1/2, x=9/4 and b=e#

#:. log_"1/2"(9/4) =ln(9/4)/ln(1/2)#

#= (ln9-ln4)/(ln1-ln2)#

#=(ln9-ln4)/(0-ln2) = - (ln9-ln4)/ln2#

#~= -1.16992#

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