Question Number 93911 by i jagooll last updated on 16/May/20 | ||
$$\mathrm{number}\:\mathrm{of}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{a}\:\mathrm{number}\: \\ $$$$\mathrm{2}^{\mathrm{2016}} \:\mathrm{and}\:\mathrm{5}^{\mathrm{2016}} \:\mathrm{is}? \\ $$ | ||
Commented by PRITHWISH SEN 2 last updated on 16/May/20 | ||
$$\mathrm{2}^{\mathrm{2016}} =\frac{\left(\mathrm{2}^{\mathrm{10}} \right)^{\mathrm{203}} }{\mathrm{2}^{\mathrm{14}} }=\frac{\left(\mathrm{1024}\right)^{\mathrm{203}} }{\mathrm{2}^{\mathrm{10}} .\mathrm{2}^{\mathrm{4}} }\:\backsim\:\frac{\left(\mathrm{10}\right)^{\mathrm{609}} }{\mathrm{10}^{\mathrm{3}} }\backsim\left(\mathrm{10}\right)^{\mathrm{606}} \\ $$$$=\mathrm{606}+\mathrm{1digits}=\mathrm{607}\:\mathrm{digits} \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{it}\:\mathrm{works}.\:\mathrm{So}\:\mathrm{please}\:\mathrm{check}. \\ $$$$\mathrm{5}^{\mathrm{2016}} =\:\frac{\mathrm{10}^{\mathrm{2016}} }{\mathrm{2}^{\mathrm{2016}} }\:=\:\frac{\mathrm{2017digits}}{\mathrm{607}\:\mathrm{digits}}\:=\:\mathrm{1410}\:\mathrm{digits} \\ $$ | ||
Commented by i jagooll last updated on 16/May/20 | ||
$$\mathrm{sir}\:\mathrm{5}^{\mathrm{2016}} \:\mathrm{i}\:\mathrm{got}\:\mathrm{1410}?\:\mathrm{it}\:\mathrm{correct}? \\ $$ | ||
Commented by Tony Lin last updated on 16/May/20 | ||
$${log}_{\mathrm{10}} \mathrm{2}^{\mathrm{2016}} =\mathrm{2016}{log}_{\mathrm{10}} \mathrm{2} \\ $$$$\approx\mathrm{606}.\mathrm{8765} \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{2016}} \approx\mathrm{10}^{\mathrm{606}.\mathrm{8765}} \approx\mathrm{7}.\mathrm{525}×\mathrm{10}^{\mathrm{606}} \\ $$$$\rightarrow\mathrm{607}\:{digits} \\ $$$${log}_{\mathrm{10}} \mathrm{5}^{\mathrm{2016}} =\mathrm{2016}{log}_{\mathrm{10}} \mathrm{5} \\ $$$$\approx\mathrm{1409}.\mathrm{1235} \\ $$$$\Rightarrow\mathrm{5}^{\mathrm{2016}} \approx\mathrm{10}^{\mathrm{1409}.\mathrm{1235}} \approx\mathrm{1}.\mathrm{329}×\mathrm{10}^{\mathrm{1409}} \\ $$$$\rightarrow\mathrm{1410}\:{digits} \\ $$ | ||
Commented by PRITHWISH SEN 2 last updated on 16/May/20 | ||
$$\mathrm{I}\:\mathrm{think}\:\mathrm{10}^{\mathrm{606}.\mathrm{8765}} =\:\mathrm{606}+\mathrm{1}=\mathrm{607digits} \\ $$$$\mathrm{and}\:\mathrm{10}^{\mathrm{1409}.\mathrm{1235}} =\mathrm{1409}+\mathrm{1}=\mathrm{1410}\:\mathrm{digits} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$ | ||
Commented by i jagooll last updated on 16/May/20 | ||
what if students don't memorize grades from log base (10) 2 and log base (10) 5 sir? | ||
Commented by i jagooll last updated on 16/May/20 | ||
$$\mathrm{is}\:\mathrm{there}\:\mathrm{another}\:\mathrm{way}? \\ $$ | ||
Commented by Tony Lin last updated on 16/May/20 | ||
$${actually} \\ $$$${log}\mathrm{5}=\mathrm{1}−{log}\mathrm{2} \\ $$$${just}\:{remember}\: \\ $$$${log}\mathrm{2}\approx\mathrm{0}.\mathrm{3010} \\ $$$${log}\mathrm{3}\approx\mathrm{0}.\mathrm{4771} \\ $$$${log}\mathrm{7}\approx\mathrm{0}.\mathrm{8451} \\ $$$${in}\:{senior}\:{high}\:{school} \\ $$$${Teachers}\:{asked}\:{us}\:{to}\:{remember}\:{these} \\ $$$${three}\:{common}\:{values},\:{which}\:{is}\:{easy} \\ $$$${to}\:{evalute}\:{a}^{\pm{b}} \:{if}\:{b}\:{is}\:{a}\:{large}\:{number} \\ $$$${And}\:{sometimes}\:{it}\:{would}\:{give}\:{you}\:{the} \\ $$$${value}\:{in}\:{the}\:{test},{so}\:{there}'{s}\:{no}\:{worry} \\ $$$${about}\:{it} \\ $$$${in}\:{university} \\ $$$${Calculators}\:{are}\:{allowed}\:{in}\:{some}\:{courses} \\ $$$${And}\:{logarithm}\:{is}\:{not}\:{the}\:{main}\:{topic} \\ $$$${of}\:{the}\:{test} \\ $$$$\:\boldsymbol{{But}} \\ $$$${One}\:{method}\:{to}\:{evaluate}\:{logarithm} \\ $$$${I}\:{came}\:{up}\:{with}\:{is}\:{using}\:{Taylor}\:{series} \\ $$$${log}\mathrm{2}=\frac{{ln}\mathrm{2}}{{ln}\mathrm{10}} \\ $$$${We}\:{know}\:{that}\: \\ $$$${ln}\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+\centerdot\centerdot\centerdot \\ $$$${for}\:−\mathrm{1}<{x}\leqslant\mathrm{1} \\ $$$${plug}\:\mathrm{1}\:{in}\: \\ $$$$\Rightarrow{ln}\mathrm{2}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\centerdot\centerdot\centerdot\approx\mathrm{0}.\mathrm{693147} \\ $$$${We}\:{cannot}\:{plug}\:\mathrm{9}\:{in}\:{the}\:{Mercator}\:{series} \\ $$$${because}\:−\mathrm{1}<{x}\leqslant\mathrm{1} \\ $$$${but}\:{we}\:{can}\:{use}\: \\ $$$${ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$$={ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right) \\ $$$${plug}\:\frac{\mathrm{9}}{\mathrm{11}}\:{in} \\ $$$${so}\:{ln}\mathrm{10} \\ $$$$=\left[\frac{\mathrm{9}}{\mathrm{11}}−\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{3}} }{\mathrm{3}}−\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{4}} }{\mathrm{4}}+\centerdot\centerdot\centerdot\right] \\ $$$$+\left[\frac{\mathrm{9}}{\mathrm{11}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{3}} }{\mathrm{3}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{4}} }{\mathrm{4}}+\centerdot\centerdot\centerdot\right] \\ $$$$=\mathrm{2}×\left[\frac{\mathrm{9}}{\mathrm{11}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{3}} }{\mathrm{3}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{5}} }{\mathrm{5}}+\centerdot\centerdot\centerdot\right] \\ $$$$=\mathrm{2}×\frac{\frac{\mathrm{9}}{\mathrm{11}}}{\mathrm{1}−\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+\centerdot\centerdot\centerdot\right) \\ $$$$\approx\mathrm{2}.\mathrm{302585} \\ $$$$\therefore{log}\mathrm{2}=\frac{{ln}\mathrm{2}}{{ln}\mathrm{10}}\approx\mathrm{0}.\mathrm{3010} \\ $$ | ||
Commented by i jagooll last updated on 16/May/20 | ||
$$\mathrm{ok}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$ | ||