Given log4=0.6021, log9=0.9542, and log12=1.-792, how do you find log 1.2?

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Answer 1 · 2017-01-08T09:57:00

#log1.2=0.0792#

As #log9=0.9542#. we have

#log9=log3^2=2log3=0.9542# i.e. #log3=0.9542/2=0.4771#

Hence #log1.2=log((4xx3)/10)=log4+log3-log10#

= #0.6021+0.4771-1#

= #1.0792-1#

= #0.0792#

We could have also done

#log1.2=log(12/10)=log12-log10#

= #1.0792-1=0.0792#