How do you calculate #antilog (0.6117 + -3) #?

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Answer 1 · 2016-10-12T15:34:25

#x=0.0040898#

Let #x=anti log(0.6117+(-3))#

then #0.6117+(-3)=logx#

or #0.6117=logx+3=log(1000x)#

and #1000x=anti log(0.6117)#

from tables, #antilog0.6117=4.0898#

and #x=0.0040898#