Show that there exists an integer that is a power of 5 with the first 4 digits being 2018.Given $2^x=129$, why is it that I can use the natural logarithm to find $x$?Satisfying equality between logarithmic expressionsDividing logarithms without using a calculatorIf $a=blog b$, how does $b$ grow asymptotically?So many logs with different basesComparing functions that have logs in exponentsHow to show that $- log_b x = log_frac1b x$Taking log of a matrix equationhow to simplify a quadratic logarithmic equation equation?
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Show that there exists an integer that is a power of 5 with the first 4 digits being 2018.
Given $2^x=129$, why is it that I can use the natural logarithm to find $x$?Satisfying equality between logarithmic expressionsDividing logarithms without using a calculatorIf $a=blog b$, how does $b$ grow asymptotically?So many logs with different basesComparing functions that have logs in exponentsHow to show that $- log_b x = log_frac1b x$Taking log of a matrix equationhow to simplify a quadratic logarithmic equation equation?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I know the problem can be expressed in the following way. Let $a,b in mathbbZ^+$ such that $2018:times:10^b < 5^a < 2019:times:10^b$
Taking logs, I get $b + log 2018 < a log 5< b + log 2019$.
Not quite sure how else to proceed though.
logarithms
$endgroup$
add a comment |
$begingroup$
I know the problem can be expressed in the following way. Let $a,b in mathbbZ^+$ such that $2018:times:10^b < 5^a < 2019:times:10^b$
Taking logs, I get $b + log 2018 < a log 5< b + log 2019$.
Not quite sure how else to proceed though.
logarithms
$endgroup$
$begingroup$
Note: $5^92=2019....$
$endgroup$
– J. W. Tanner
10 hours ago
5
$begingroup$
$5^4364=2.018826153906575ldotscdot 10^3050$.
$endgroup$
– Christian Blatter
10 hours ago
add a comment |
$begingroup$
I know the problem can be expressed in the following way. Let $a,b in mathbbZ^+$ such that $2018:times:10^b < 5^a < 2019:times:10^b$
Taking logs, I get $b + log 2018 < a log 5< b + log 2019$.
Not quite sure how else to proceed though.
logarithms
$endgroup$
I know the problem can be expressed in the following way. Let $a,b in mathbbZ^+$ such that $2018:times:10^b < 5^a < 2019:times:10^b$
Taking logs, I get $b + log 2018 < a log 5< b + log 2019$.
Not quite sure how else to proceed though.
logarithms
logarithms
asked 10 hours ago
Haikal YeoHaikal Yeo
1,0638 silver badges23 bronze badges
1,0638 silver badges23 bronze badges
$begingroup$
Note: $5^92=2019....$
$endgroup$
– J. W. Tanner
10 hours ago
5
$begingroup$
$5^4364=2.018826153906575ldotscdot 10^3050$.
$endgroup$
– Christian Blatter
10 hours ago
add a comment |
$begingroup$
Note: $5^92=2019....$
$endgroup$
– J. W. Tanner
10 hours ago
5
$begingroup$
$5^4364=2.018826153906575ldotscdot 10^3050$.
$endgroup$
– Christian Blatter
10 hours ago
$begingroup$
Note: $5^92=2019....$
$endgroup$
– J. W. Tanner
10 hours ago
$begingroup$
Note: $5^92=2019....$
$endgroup$
– J. W. Tanner
10 hours ago
5
5
$begingroup$
$5^4364=2.018826153906575ldotscdot 10^3050$.
$endgroup$
– Christian Blatter
10 hours ago
$begingroup$
$5^4364=2.018826153906575ldotscdot 10^3050$.
$endgroup$
– Christian Blatter
10 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The set $A = a log_10 5 + b : a,b in mathbb Z $ is an additive subgroup of $mathbb R$, and so is either cyclic or dense.
If $A$ were cyclic, then $1 = n (a_0 log_10 5 + b_0)$. Since $log_10 5$ is irrational, we must have $n a_0 =0$ and $n b_0 = 1$, which cannot happen.
Therefore, $A$ is dense and so there is an element of $A$ in the interval $(log_10(2018), log_10(2019))$.
$endgroup$
$begingroup$
But what if $a <0$? ;)
$endgroup$
– N. S.
20 mins ago
add a comment |
$begingroup$
Hint Since $log_10(5)$ is irrational, the set $ a log_10(5)-b : a, b in mathbb N$ is dense in $mathbb R$. Therefore, you can find such a number in the interval $(log_10(2018), log_10(2019))$.
$endgroup$
add a comment |
$begingroup$
201882615390657491958528733888359303897206738410754045263963565695705442119457224168899440239407953138581226299365782567442925725493850667127823188129261463991440173793523996779651936654389368686597194439736208695989862156299629602332523464076512669290731774513558552819180354651303963581641490360308535552425419867299337653389070510045424468908107895566363643482640162372352948553283421340038905458554369108404396575106064121414101960663718546239386527523409161552124792218268239338974867937280489575854266372435173446464066145926791484234116593481699639727998774125414871721332982942016722814109084246332819815465119939678154783718052333181096038896636579784807212135999682794464880561777058643610122484366550692964046735317370672168782581592297953410438100999375161197662519910850059633016528523568042285302742062098814548246723733888013653327016324124727737686998618632277442820374655493093729902407403246796933589232379520684294387357381354286158732655143619009740514672918111683766211391182403531198532559980151972513489907217185072196731637657634674068559899298805500983571984624086190968818017889989749154302003358764408667797989976730972834651878998380425225936897264605310688663029535576600866502864768965770091351971521995641025227860905919121596849003861094298243892705486363508792818416683489930886591576035182909921868555205172091169698191585767546957494046847024226730833257159854651007706945782566951387958682409697454133730086314492661919431019883237423708525758459729006177183221902634043244036721957840616924771617991861189108596959612503053521704057188513232970211432467080796220044513875134615739794385874670476886205780889353138817697126168129565683692421875294802472389184825853664308097076451463513926649779788130377406299829608329206810680213500593258029567966685039842829425254271550324266936497912430799186821435764991658842175752107255972699604608610838678647922991780897811201255157892448118302260686546980422098465087256649477365011881596878199059608414801013417027677837816041537974933022551966344978889501537172467203506670958637174099226979184380737481324329743693108262765896755437598412730342468865474573477004619362456714144985785078817902129543526158982694832379771931581941839225962375548217493133113050127913932315166413560509241349541568784192095222792848694381349868462030138524690958834727087880318746593234222439280165442150519813234371751409660250721132893912068388675454902832327537395026029852373100586738988362442544910516664069691179659777362156805235491174484504759330371215949360906031780262687660002512614551478392076625118081138080259140457229066530352113989013066692570397437773702413957231955020804389634415750498870634142816046788045672568297485242527508630207469164390742003932759774267358556602879306918807555750836800249985550904087271797537215843856934114642382854402532562504961847107385321625601211533133161511192639301728105746220687191517987404378034935607228167440448968116089808516638097060020070846318846918229693351291405338045145821917930096557614039198824684717692434787750244140625 = 5 ^ 4346
New contributor
$endgroup$
2
$begingroup$
While an exact answer does answer the question, exactly how you worked that out would probably be more useful to the person asking the question.
$endgroup$
– Ispil
38 mins ago
add a comment |
Your Answer
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3 Answers
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3 Answers
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$begingroup$
The set $A = a log_10 5 + b : a,b in mathbb Z $ is an additive subgroup of $mathbb R$, and so is either cyclic or dense.
If $A$ were cyclic, then $1 = n (a_0 log_10 5 + b_0)$. Since $log_10 5$ is irrational, we must have $n a_0 =0$ and $n b_0 = 1$, which cannot happen.
Therefore, $A$ is dense and so there is an element of $A$ in the interval $(log_10(2018), log_10(2019))$.
$endgroup$
$begingroup$
But what if $a <0$? ;)
$endgroup$
– N. S.
20 mins ago
add a comment |
$begingroup$
The set $A = a log_10 5 + b : a,b in mathbb Z $ is an additive subgroup of $mathbb R$, and so is either cyclic or dense.
If $A$ were cyclic, then $1 = n (a_0 log_10 5 + b_0)$. Since $log_10 5$ is irrational, we must have $n a_0 =0$ and $n b_0 = 1$, which cannot happen.
Therefore, $A$ is dense and so there is an element of $A$ in the interval $(log_10(2018), log_10(2019))$.
$endgroup$
$begingroup$
But what if $a <0$? ;)
$endgroup$
– N. S.
20 mins ago
add a comment |
$begingroup$
The set $A = a log_10 5 + b : a,b in mathbb Z $ is an additive subgroup of $mathbb R$, and so is either cyclic or dense.
If $A$ were cyclic, then $1 = n (a_0 log_10 5 + b_0)$. Since $log_10 5$ is irrational, we must have $n a_0 =0$ and $n b_0 = 1$, which cannot happen.
Therefore, $A$ is dense and so there is an element of $A$ in the interval $(log_10(2018), log_10(2019))$.
$endgroup$
The set $A = a log_10 5 + b : a,b in mathbb Z $ is an additive subgroup of $mathbb R$, and so is either cyclic or dense.
If $A$ were cyclic, then $1 = n (a_0 log_10 5 + b_0)$. Since $log_10 5$ is irrational, we must have $n a_0 =0$ and $n b_0 = 1$, which cannot happen.
Therefore, $A$ is dense and so there is an element of $A$ in the interval $(log_10(2018), log_10(2019))$.
answered 9 hours ago
lhflhf
174k12 gold badges179 silver badges418 bronze badges
174k12 gold badges179 silver badges418 bronze badges
$begingroup$
But what if $a <0$? ;)
$endgroup$
– N. S.
20 mins ago
add a comment |
$begingroup$
But what if $a <0$? ;)
$endgroup$
– N. S.
20 mins ago
$begingroup$
But what if $a <0$? ;)
$endgroup$
– N. S.
20 mins ago
$begingroup$
But what if $a <0$? ;)
$endgroup$
– N. S.
20 mins ago
add a comment |
$begingroup$
Hint Since $log_10(5)$ is irrational, the set $ a log_10(5)-b : a, b in mathbb N$ is dense in $mathbb R$. Therefore, you can find such a number in the interval $(log_10(2018), log_10(2019))$.
$endgroup$
add a comment |
$begingroup$
Hint Since $log_10(5)$ is irrational, the set $ a log_10(5)-b : a, b in mathbb N$ is dense in $mathbb R$. Therefore, you can find such a number in the interval $(log_10(2018), log_10(2019))$.
$endgroup$
add a comment |
$begingroup$
Hint Since $log_10(5)$ is irrational, the set $ a log_10(5)-b : a, b in mathbb N$ is dense in $mathbb R$. Therefore, you can find such a number in the interval $(log_10(2018), log_10(2019))$.
$endgroup$
Hint Since $log_10(5)$ is irrational, the set $ a log_10(5)-b : a, b in mathbb N$ is dense in $mathbb R$. Therefore, you can find such a number in the interval $(log_10(2018), log_10(2019))$.
answered 10 hours ago
N. S.N. S.
109k7 gold badges118 silver badges215 bronze badges
109k7 gold badges118 silver badges215 bronze badges
add a comment |
add a comment |
$begingroup$
201882615390657491958528733888359303897206738410754045263963565695705442119457224168899440239407953138581226299365782567442925725493850667127823188129261463991440173793523996779651936654389368686597194439736208695989862156299629602332523464076512669290731774513558552819180354651303963581641490360308535552425419867299337653389070510045424468908107895566363643482640162372352948553283421340038905458554369108404396575106064121414101960663718546239386527523409161552124792218268239338974867937280489575854266372435173446464066145926791484234116593481699639727998774125414871721332982942016722814109084246332819815465119939678154783718052333181096038896636579784807212135999682794464880561777058643610122484366550692964046735317370672168782581592297953410438100999375161197662519910850059633016528523568042285302742062098814548246723733888013653327016324124727737686998618632277442820374655493093729902407403246796933589232379520684294387357381354286158732655143619009740514672918111683766211391182403531198532559980151972513489907217185072196731637657634674068559899298805500983571984624086190968818017889989749154302003358764408667797989976730972834651878998380425225936897264605310688663029535576600866502864768965770091351971521995641025227860905919121596849003861094298243892705486363508792818416683489930886591576035182909921868555205172091169698191585767546957494046847024226730833257159854651007706945782566951387958682409697454133730086314492661919431019883237423708525758459729006177183221902634043244036721957840616924771617991861189108596959612503053521704057188513232970211432467080796220044513875134615739794385874670476886205780889353138817697126168129565683692421875294802472389184825853664308097076451463513926649779788130377406299829608329206810680213500593258029567966685039842829425254271550324266936497912430799186821435764991658842175752107255972699604608610838678647922991780897811201255157892448118302260686546980422098465087256649477365011881596878199059608414801013417027677837816041537974933022551966344978889501537172467203506670958637174099226979184380737481324329743693108262765896755437598412730342468865474573477004619362456714144985785078817902129543526158982694832379771931581941839225962375548217493133113050127913932315166413560509241349541568784192095222792848694381349868462030138524690958834727087880318746593234222439280165442150519813234371751409660250721132893912068388675454902832327537395026029852373100586738988362442544910516664069691179659777362156805235491174484504759330371215949360906031780262687660002512614551478392076625118081138080259140457229066530352113989013066692570397437773702413957231955020804389634415750498870634142816046788045672568297485242527508630207469164390742003932759774267358556602879306918807555750836800249985550904087271797537215843856934114642382854402532562504961847107385321625601211533133161511192639301728105746220687191517987404378034935607228167440448968116089808516638097060020070846318846918229693351291405338045145821917930096557614039198824684717692434787750244140625 = 5 ^ 4346
New contributor
$endgroup$
2
$begingroup$
While an exact answer does answer the question, exactly how you worked that out would probably be more useful to the person asking the question.
$endgroup$
– Ispil
38 mins ago
add a comment |
$begingroup$
201882615390657491958528733888359303897206738410754045263963565695705442119457224168899440239407953138581226299365782567442925725493850667127823188129261463991440173793523996779651936654389368686597194439736208695989862156299629602332523464076512669290731774513558552819180354651303963581641490360308535552425419867299337653389070510045424468908107895566363643482640162372352948553283421340038905458554369108404396575106064121414101960663718546239386527523409161552124792218268239338974867937280489575854266372435173446464066145926791484234116593481699639727998774125414871721332982942016722814109084246332819815465119939678154783718052333181096038896636579784807212135999682794464880561777058643610122484366550692964046735317370672168782581592297953410438100999375161197662519910850059633016528523568042285302742062098814548246723733888013653327016324124727737686998618632277442820374655493093729902407403246796933589232379520684294387357381354286158732655143619009740514672918111683766211391182403531198532559980151972513489907217185072196731637657634674068559899298805500983571984624086190968818017889989749154302003358764408667797989976730972834651878998380425225936897264605310688663029535576600866502864768965770091351971521995641025227860905919121596849003861094298243892705486363508792818416683489930886591576035182909921868555205172091169698191585767546957494046847024226730833257159854651007706945782566951387958682409697454133730086314492661919431019883237423708525758459729006177183221902634043244036721957840616924771617991861189108596959612503053521704057188513232970211432467080796220044513875134615739794385874670476886205780889353138817697126168129565683692421875294802472389184825853664308097076451463513926649779788130377406299829608329206810680213500593258029567966685039842829425254271550324266936497912430799186821435764991658842175752107255972699604608610838678647922991780897811201255157892448118302260686546980422098465087256649477365011881596878199059608414801013417027677837816041537974933022551966344978889501537172467203506670958637174099226979184380737481324329743693108262765896755437598412730342468865474573477004619362456714144985785078817902129543526158982694832379771931581941839225962375548217493133113050127913932315166413560509241349541568784192095222792848694381349868462030138524690958834727087880318746593234222439280165442150519813234371751409660250721132893912068388675454902832327537395026029852373100586738988362442544910516664069691179659777362156805235491174484504759330371215949360906031780262687660002512614551478392076625118081138080259140457229066530352113989013066692570397437773702413957231955020804389634415750498870634142816046788045672568297485242527508630207469164390742003932759774267358556602879306918807555750836800249985550904087271797537215843856934114642382854402532562504961847107385321625601211533133161511192639301728105746220687191517987404378034935607228167440448968116089808516638097060020070846318846918229693351291405338045145821917930096557614039198824684717692434787750244140625 = 5 ^ 4346
New contributor
$endgroup$
2
$begingroup$
While an exact answer does answer the question, exactly how you worked that out would probably be more useful to the person asking the question.
$endgroup$
– Ispil
38 mins ago
add a comment |
$begingroup$
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New contributor
$endgroup$
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edited 16 mins ago
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answered 49 mins ago
andrew.punnettandrew.punnett
933 bronze badges
933 bronze badges
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2
$begingroup$
While an exact answer does answer the question, exactly how you worked that out would probably be more useful to the person asking the question.
$endgroup$
– Ispil
38 mins ago
add a comment |
2
$begingroup$
While an exact answer does answer the question, exactly how you worked that out would probably be more useful to the person asking the question.
$endgroup$
– Ispil
38 mins ago
2
2
$begingroup$
While an exact answer does answer the question, exactly how you worked that out would probably be more useful to the person asking the question.
$endgroup$
– Ispil
38 mins ago
$begingroup$
While an exact answer does answer the question, exactly how you worked that out would probably be more useful to the person asking the question.
$endgroup$
– Ispil
38 mins ago
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$begingroup$
Note: $5^92=2019....$
$endgroup$
– J. W. Tanner
10 hours ago
5
$begingroup$
$5^4364=2.018826153906575ldotscdot 10^3050$.
$endgroup$
– Christian Blatter
10 hours ago