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Question Number 93911 by i jagooll last updated on 16/May/20
number of digit of a number   2^(2016)  and 5^(2016)  is?
$$\mathrm{number}\:\mathrm{of}\:\mathrm{digit}\:\mathrm{of}\:\mathrm{a}\:\mathrm{number}\: \\ $$$$\mathrm{2}^{\mathrm{2016}} \:\mathrm{and}\:\mathrm{5}^{\mathrm{2016}} \:\mathrm{is}? \\ $$
Commented by PRITHWISH SEN 2 last updated on 16/May/20
2^(2016) =(((2^(10) )^(203) )/2^(14) )=(((1024)^(203) )/(2^(10) .2^4 )) ∽ (((10)^(609) )/(10^3 ))∽(10)^(606)   =606+1digits=607 digits  I am not sure if it works. So please check.  5^(2016) = ((10^(2016) )/2^(2016) ) = ((2017digits)/(607 digits)) = 1410 digits
$$\mathrm{2}^{\mathrm{2016}} =\frac{\left(\mathrm{2}^{\mathrm{10}} \right)^{\mathrm{203}} }{\mathrm{2}^{\mathrm{14}} }=\frac{\left(\mathrm{1024}\right)^{\mathrm{203}} }{\mathrm{2}^{\mathrm{10}} .\mathrm{2}^{\mathrm{4}} }\:\backsim\:\frac{\left(\mathrm{10}\right)^{\mathrm{609}} }{\mathrm{10}^{\mathrm{3}} }\backsim\left(\mathrm{10}\right)^{\mathrm{606}} \\ $$$$=\mathrm{606}+\mathrm{1digits}=\mathrm{607}\:\mathrm{digits} \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{it}\:\mathrm{works}.\:\mathrm{So}\:\mathrm{please}\:\mathrm{check}. \\ $$$$\mathrm{5}^{\mathrm{2016}} =\:\frac{\mathrm{10}^{\mathrm{2016}} }{\mathrm{2}^{\mathrm{2016}} }\:=\:\frac{\mathrm{2017digits}}{\mathrm{607}\:\mathrm{digits}}\:=\:\mathrm{1410}\:\mathrm{digits} \\ $$
Commented by i jagooll last updated on 16/May/20
sir 5^(2016)  i got 1410? it correct?
$$\mathrm{sir}\:\mathrm{5}^{\mathrm{2016}} \:\mathrm{i}\:\mathrm{got}\:\mathrm{1410}?\:\mathrm{it}\:\mathrm{correct}? \\ $$
Commented by Tony Lin last updated on 16/May/20
log_(10) 2^(2016) =2016log_(10) 2  ≈606.8765  ⇒2^(2016) ≈10^(606.8765) ≈7.525×10^(606)   →607 digits  log_(10) 5^(2016) =2016log_(10) 5  ≈1409.1235  ⇒5^(2016) ≈10^(1409.1235) ≈1.329×10^(1409)   →1410 digits
$${log}_{\mathrm{10}} \mathrm{2}^{\mathrm{2016}} =\mathrm{2016}{log}_{\mathrm{10}} \mathrm{2} \\ $$$$\approx\mathrm{606}.\mathrm{8765} \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{2016}} \approx\mathrm{10}^{\mathrm{606}.\mathrm{8765}} \approx\mathrm{7}.\mathrm{525}×\mathrm{10}^{\mathrm{606}} \\ $$$$\rightarrow\mathrm{607}\:{digits} \\ $$$${log}_{\mathrm{10}} \mathrm{5}^{\mathrm{2016}} =\mathrm{2016}{log}_{\mathrm{10}} \mathrm{5} \\ $$$$\approx\mathrm{1409}.\mathrm{1235} \\ $$$$\Rightarrow\mathrm{5}^{\mathrm{2016}} \approx\mathrm{10}^{\mathrm{1409}.\mathrm{1235}} \approx\mathrm{1}.\mathrm{329}×\mathrm{10}^{\mathrm{1409}} \\ $$$$\rightarrow\mathrm{1410}\:{digits} \\ $$
Commented by PRITHWISH SEN 2 last updated on 16/May/20
I think 10^(606.8765) = 606+1=607digits  and 10^(1409.1235) =1409+1=1410 digits  thank you sir.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{10}^{\mathrm{606}.\mathrm{8765}} =\:\mathrm{606}+\mathrm{1}=\mathrm{607digits} \\ $$$$\mathrm{and}\:\mathrm{10}^{\mathrm{1409}.\mathrm{1235}} =\mathrm{1409}+\mathrm{1}=\mathrm{1410}\:\mathrm{digits} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
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
is there another way?
$$\mathrm{is}\:\mathrm{there}\:\mathrm{another}\:\mathrm{way}? \\ $$
Commented by Tony Lin last updated on 16/May/20
actually  log5=1−log2  just remember   log2≈0.3010  log3≈0.4771  log7≈0.8451  in senior high school  Teachers asked us to remember these  three common values, which is easy  to evalute a^(±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   But  One method to evaluate logarithm  I came up with is using Taylor series  log2=((ln2)/(ln10))  We know that   ln(1+x)=x−(x^2 /2)+(x^3 /3)−(x^4 /4)+∙∙∙  for −1<x≤1  plug 1 in   ⇒ln2=1−(1/2)+(1/3)−(1/4)+∙∙∙≈0.693147  We cannot plug 9 in the Mercator series  because −1<x≤1  but we can use   ln(((1+x)/(1−x)))  =ln(1+x)−ln(1−x)  plug (9/(11)) in  so ln10  =[(9/(11))−((((9/(11)))^2 )/2)+((((9/(11)))^3 )/3)−((((9/(11)))^4 )/4)+∙∙∙]  +[(9/(11))+((((9/(11)))^2 )/2)+((((9/(11)))^3 )/3)+((((9/(11)))^4 )/4)+∙∙∙]  =2×[(9/(11))+((((9/(11)))^3 )/3)+((((9/(11)))^5 )/5)+∙∙∙]  =2×((9/(11))/(1−((9/(11)))^2 ))(1+(1/3)+(1/5)+∙∙∙)  ≈2.302585  ∴log2=((ln2)/(ln10))≈0.3010
$${actually} \\ $$$${log}\mathrm{5}=\mathrm{1}−{log}\mathrm{2} \\ $$$${just}\:{remember}\: \\ $$$${log}\mathrm{2}\approx\mathrm{0}.\mathrm{3010} \\ $$$${log}\mathrm{3}\approx\mathrm{0}.\mathrm{4771} \\ $$$${log}\mathrm{7}\approx\mathrm{0}.\mathrm{8451} \\ $$$${in}\:{senior}\:{high}\:{school} \\ $$$${Teachers}\:{asked}\:{us}\:{to}\:{remember}\:{these} \\ $$$${three}\:{common}\:{values},\:{which}\:{is}\:{easy} \\ $$$${to}\:{evalute}\:{a}^{\pm{b}} \:{if}\:{b}\:{is}\:{a}\:{large}\:{number} \\ $$$${And}\:{sometimes}\:{it}\:{would}\:{give}\:{you}\:{the} \\ $$$${value}\:{in}\:{the}\:{test},{so}\:{there}'{s}\:{no}\:{worry} \\ $$$${about}\:{it} \\ $$$${in}\:{university} \\ $$$${Calculators}\:{are}\:{allowed}\:{in}\:{some}\:{courses} \\ $$$${And}\:{logarithm}\:{is}\:{not}\:{the}\:{main}\:{topic} \\ $$$${of}\:{the}\:{test} \\ $$$$\:\boldsymbol{{But}} \\ $$$${One}\:{method}\:{to}\:{evaluate}\:{logarithm} \\ $$$${I}\:{came}\:{up}\:{with}\:{is}\:{using}\:{Taylor}\:{series} \\ $$$${log}\mathrm{2}=\frac{{ln}\mathrm{2}}{{ln}\mathrm{10}} \\ $$$${We}\:{know}\:{that}\: \\ $$$${ln}\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+\centerdot\centerdot\centerdot \\ $$$${for}\:−\mathrm{1}<{x}\leqslant\mathrm{1} \\ $$$${plug}\:\mathrm{1}\:{in}\: \\ $$$$\Rightarrow{ln}\mathrm{2}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+\centerdot\centerdot\centerdot\approx\mathrm{0}.\mathrm{693147} \\ $$$${We}\:{cannot}\:{plug}\:\mathrm{9}\:{in}\:{the}\:{Mercator}\:{series} \\ $$$${because}\:−\mathrm{1}<{x}\leqslant\mathrm{1} \\ $$$${but}\:{we}\:{can}\:{use}\: \\ $$$${ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$$={ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{1}−{x}\right) \\ $$$${plug}\:\frac{\mathrm{9}}{\mathrm{11}}\:{in} \\ $$$${so}\:{ln}\mathrm{10} \\ $$$$=\left[\frac{\mathrm{9}}{\mathrm{11}}−\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{3}} }{\mathrm{3}}−\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{4}} }{\mathrm{4}}+\centerdot\centerdot\centerdot\right] \\ $$$$+\left[\frac{\mathrm{9}}{\mathrm{11}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{3}} }{\mathrm{3}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{4}} }{\mathrm{4}}+\centerdot\centerdot\centerdot\right] \\ $$$$=\mathrm{2}×\left[\frac{\mathrm{9}}{\mathrm{11}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{3}} }{\mathrm{3}}+\frac{\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{5}} }{\mathrm{5}}+\centerdot\centerdot\centerdot\right] \\ $$$$=\mathrm{2}×\frac{\frac{\mathrm{9}}{\mathrm{11}}}{\mathrm{1}−\left(\frac{\mathrm{9}}{\mathrm{11}}\right)^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+\centerdot\centerdot\centerdot\right) \\ $$$$\approx\mathrm{2}.\mathrm{302585} \\ $$$$\therefore{log}\mathrm{2}=\frac{{ln}\mathrm{2}}{{ln}\mathrm{10}}\approx\mathrm{0}.\mathrm{3010} \\ $$
Commented by i jagooll last updated on 16/May/20
ok sir. thank you
$$\mathrm{ok}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$

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