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

$log 1.2 = 0.0792$ $log1.2=0.0792$

As $log 9 = 0.9542$ $log9=0.9542$ . we have

$log 9 = {log 3}^{2} = 2 log 3 = 0.9542$ $log9=log3^2=2log3=0.9542$ i.e. $log 3 = \frac{0.9542}{2} = 0.4771$ $log3=0.9542/2=0.4771$

Hence $log 1.2 = log (\frac{4 \times 3}{10}) = log 4 + log 3 - log 10$ $log1.2=log((4xx3)/10)=log4+log3-log10$

= $0.6021 + 0.4771 - 1$ $0.6021+0.4771-1$

= $1.0792 - 1$ $1.0792-1$

= $0.0792$ $0.0792$

We could have also done

$log 1.2 = log (\frac{12}{10}) = log 12 - log 10$ $log1.2=log(12/10)=log12-log10$

= $1.0792 - 1 = 0.0792$ $1.0792-1=0.0792$