Menu Close

Having-given-log-2-0-30103-find-the-position-of-the-first-significant-figure-in-2-37-

Tinku Tara 0 Comments
Pin




Question Number 58060 by Kunal12588 last updated on 17/Apr/19
Having given log 2 = 0.30103, find the position of the first significant figure in 2^(−37) .
$${Having}\:{given}\:\mathrm{log}\:\mathrm{2}\:=\:\mathrm{0}.\mathrm{30103},\:{find}\:{the}\:{position} \\ $$$${of}\:{the}\:{first}\:{significant}\:{figure}\:{in}\:\mathrm{2}^{−\mathrm{37}} . \\ $$
Commented by Rasheed.Sindhi last updated on 17/Apr/19
−11.13811≠11^(−) .13811 −11.13811 is whole negative i-e it is equal to −11−.13811 While in 11^(−) .13811 only integer part is negative. I-e 11^(−) .13811=−11+.13811 As −11.13811 is whole negative, therefore in it .13811 is not mantissa (and −11 is also not characteristics.) because mantissa is always positive To obtain mantissa and characteristics from −11.13811 thefollowing process is mentioned: −11.13811=−11.13811+12−12 =(−11.13811+12)−12=.86189−12 =12^(−) .86189
$$−\mathrm{11}.\mathrm{13811}\neq\overline {\mathrm{11}}.\mathrm{13811} \\ $$$$−\mathrm{11}.\mathrm{13811}\:{is}\:{whole}\:{negative} \\ $$$${i}-{e}\:{it}\:{is}\:{equal}\:{to}\:\:−\mathrm{11}−.\mathrm{13811} \\ $$$${While}\:{in}\:\overline {\mathrm{11}}.\mathrm{13811}\:{only}\:{integer}\:{part}\:{is} \\ $$$${negative}.\:{I}-{e}\:\overline {\mathrm{11}}.\mathrm{13811}=−\mathrm{11}+.\mathrm{13811} \\ $$$${As}\:−\mathrm{11}.\mathrm{13811}\:{is}\:{whole}\:{negative}, \\ $$$${therefore}\:{in}\:{it}\:.\mathrm{13811}\:{is}\:{not}\:{mantissa} \\ $$$$\left({and}\:−\mathrm{11}\:{is}\:{also}\:{not}\:{characteristics}.\right) \\ $$$${because}\:{mantissa}\:{is}\:{always}\:{positive} \\ $$$${To}\:{obtain}\:{mantissa}\:{and}\:{characteristics} \\ $$$${from}\:−\mathrm{11}.\mathrm{13811}\:{thefollowing}\:{process} \\ $$$${is}\:{mentioned}: \\ $$$$\:\:\:\:\:−\mathrm{11}.\mathrm{13811}=−\mathrm{11}.\mathrm{13811}+\mathrm{12}−\mathrm{12} \\ $$$$=\left(−\mathrm{11}.\mathrm{13811}+\mathrm{12}\right)−\mathrm{12}=.\mathrm{86189}−\mathrm{12} \\ $$$$=\overline {\mathrm{12}}.\mathrm{86189} \\ $$
Commented by Kunal12588 last updated on 17/Apr/19
ohh... I got it sir thanks a lot.
$${ohh}…\:{I}\:{got}\:{it}\:{sir}\:{thanks}\:{a}\:{lot}. \\ $$
Commented by Kunal12588 last updated on 17/Apr/19
Answer given in the book :− log 2^(−37) =−37 log 2 = −37 × 0.30103 =−11.13811=12^(−) .86189 Hence in 2^(−37) there are 11 cyphers following the decimal point, i.e. the first significant figure is in the twelfth place of decimals. My Question is why and how they have done the last line. −11.13811=11^(−) .13811=12^(−) .86189
$${Answer}\:{given}\:{in}\:{the}\:{book}\::− \\ $$$$\mathrm{log}\:\:\mathrm{2}^{−\mathrm{37}} =−\mathrm{37}\:\mathrm{log}\:\mathrm{2}\:=\:−\mathrm{37}\:×\:\mathrm{0}.\mathrm{30103} \\ $$$$=−\mathrm{11}.\mathrm{13811}=\overline {\mathrm{12}}.\mathrm{86189} \\ $$$${Hence}\:{in}\:\mathrm{2}^{−\mathrm{37}} \:{there}\:{are}\:\mathrm{11}\:{cyphers}\:{following} \\ $$$${the}\:{decimal}\:{point},\:{i}.{e}.\:{the}\:{first}\:{significant} \\ $$$${figure}\:{is}\:{in}\:{the}\:{twelfth}\:{place}\:{of}\:{decimals}. \\ $$$${My}\:{Question}\:{is}\:{why}\:\:{and}\:{how}\:{they}\:{have}\:{done} \\ $$$${the}\:{last}\:{line}.\:−\mathrm{11}.\mathrm{13811}=\overline {\mathrm{11}}.\mathrm{13811}=\overline {\mathrm{12}}.\mathrm{86189} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *