Having given log 2 = 0.30103, find the position of the first significant figure in 2^(−37) .


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Question Number 58060 by Kunal12588 last updated on 17/Apr/19

$H a v i n g g i v e n \log 2 = 0.30103, f i n d t h e p o s i t i o n$ $o f t h e f i r s t s i g n i f i c a n t f i g u r e i n 2^{- 37} .$
Commented by Rasheed.Sindhi last updated on 17/Apr/19

$- 11.13811 \neq \overset{―}{11} . 13811$ $- 11.13811 i s w h o l e n e g a t i v e$ $i - e i t i s e q u a l t o - 11 - . 13811$ $W h i l e i n \overset{―}{11} . 13811 o n l y i n t e g e r p a r t i s$ $n e g a t i v e . I - e \overset{―}{11} . 13811 = - 11 + . 13811$ $A s - 11.13811 i s w h o l e n e g a t i v e,$ $t h e r e f o r e i n i t . 13811 i s n o t m a n t i s s a$ $(a n d - 11 i s a l s o n o t c h a r a c t e r i s t i c s .)$ $b e c a u s e m a n t i s s a i s a l w a y s p o s i t i v e$ $T o o b t a i n m a n t i s s a a n d c h a r a c t e r i s t i c s$ $f r o m - 11.13811 t h e f o l l o w i n g p r o c e s s$ $i s m e n t i o n e d :$ $- 11.13811 = - 11.13811 + 12 - 12$ $= (- 11.13811 + 12) - 12 = . 86189 - 12$ $= \overset{―}{12} . 86189$
Commented by Kunal12588 last updated on 17/Apr/19

$o h h . . . I g o t i t s i r t h a n k s a l o t .$
Commented by Kunal12588 last updated on 17/Apr/19

$A n s w e r g i v e n i n t h e b o o k : -$ $\log 2^{- 37} = - 37 \log 2 = - 37 \times 0.30103$ $= - 11.13811 = \overset{―}{12} . 86189$ $H e n c e i n 2^{- 37} t h e r e a r e 11 c y p h e r s f o l l o w i n g$ $t h e d e c i m a l p o i n t, i . e . t h e f i r s t s i g n i f i c a n t$ $f i g u r e i s i n t h e t w e l f t h p l a c e o f d e c i m a l s .$ $M y Q u e s t i o n i s w h y a n d h o w t h e y h a v e d o n e$ $t h e l a s t l i n e . - 11.13811 = \overset{―}{11} . 13811 = \overset{―}{12} . 86189$
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