sign up I login
 advanced
refer a friend - earn nickels!!

Ask & Discuss Questions with Community & Experts

 Moderation Team
 90 chars left    advanced
Ask iit jee aieee pet cbse icse state board community Community Discussion Question: method? JUST TRY..?2000^2001 or 2001^2000 which is greater???????
Forum Index -> Algebra like the article? email it to a friend.  
Author Message
[Post New]16 Jul 2008 17:52:52 IST
    Subject: method? JUST TRY..?2000^2001 or 2001^2000 which is greater???????
[Up] View Post
kimi (5)

Cool goIITian

Olaaa!! Perrrfect answer. 1  [1 rates]
kimi's Avatar
total posts: 34    
offline Offline
method? JUST TRY..?2000^2001 or 2001^2000 which is greater???????
    
[Post New]16 Jul 2008 18:02:43 IST
    Subject: Re:method??2000^2001 or 2001^2000 which is greater???????
[Up] View Post
krish1092 (528)

Blazing goIITian

Olaaa!! Perrrfect answer. 84  [138 rates]
krish1092's Avatar
total posts: 343    
offline Offline

Let x=20002001                         y=20012000


apply log both sides                    apply log both sides  


logx=2001(log(2000))                  logy=2000(log2001)


logx=2001(log(2*1000))                logy=2000(3.3012)


logx=2001(log2+log1000)              logy=6602.4  


logx=2001(0.3010+3)


logx=2001(3.3010)


logx=6605.301


Therefore logx > logy


                  x > y


           20002001 > 20012000




 


Rate me if useful
 this reply: 0 points  (with 0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
[Post New]16 Jul 2008 18:04:39 IST
    Subject: method? JUST TRY..?2000^2001 or 2001^2000 which is greater???????
Accepted Answer [?] [Up] View Post
rudra.panda (2559)

Blazing goIITian

Olaaa!! Perrrfect answer. 437  [624 rates]
rudra.panda's Avatar
total posts: 2533    
offline Offline
I think you can also use congruency for this.
God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
 this reply: 0 points  (with 0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
[Post New]16 Jul 2008 19:01:52 IST
    Subject: method? JUST TRY..?2000^2001 or 2001^2000 which is greater???????
[Up] View Post
kimi (5)

Cool goIITian

Olaaa!! Perrrfect answer. 1  [1 rates]
kimi's Avatar
total posts: 34    
offline Offline
how congureny???????/
 this reply: 0 points  (with 0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
[Post New]16 Jul 2008 19:11:43 IST
    Subject: Re:method? JUST TRY..?2000^2001 or 2001^2000 which is greater???????
[Up] View Post
oneyeartogo (217)

Hot goIITian

Olaaa!! Perrrfect answer. 39  [50 rates]
oneyeartogo's Avatar
total posts: 180    
offline Offline

You can try proving it using LMVT (lagrange man value theorem)
 this reply: 0 points  (with 0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
[Post New]16 Jul 2008 19:35:09 IST
    Subject: Re:method? JUST TRY..?2000^2001 or 2001^2000 which is greater???????
[Up] View Post
sachinguptaiit (940)

Scorching goIITian

Olaaa!! Perrrfect answer. 166  [221 rates]
sachinguptaiit's Avatar
total posts: 203    
offline Offline

\text{Let, }(2000)^{2001}>(2001)^{2000}\\\\\text{taking log both sides}\\\\2001\log(2000)>2000\log(2001)\\\\(2000+1)\log(2000)>2000\log(2001)\\\\\log(2000)>2000\log\left(\frac{2001}{2000}\right)\\\\\left(\frac{1}{2000}\right).\log(2000)>\log\left(1+\frac{1}{2000}\right)\\\\\text{Since, }x\ge ln(1+x)\;\forall\;x\text{ belongs to }\mathbf{R}\\\\\text{So,we can say }\left(\frac{1}{2000}\right)\ge \log\left(1+\frac{1}{2000}\right)\\\\\text{Hence, }\left(\frac{1}{2000}\right).\log(2000)>\log\left(1+\frac{1}{2000}\right)\\\\\text{Since,above is true our assumption is also true}


 


 
"Before you start some work, always ask yourself three questions - Why am I doing it, What the results might be and Will I be successful. Only when you think deeply and find satisfactory answers to these questions, go ahead."

Chanakya quotes (Indian politician, strategist and writer, 350 BC-275 BC)
 this reply: 0 points  (with 0 Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
[Post New]16 Jul 2008 19:45:15 IST
    Subject: Re:method? JUST TRY..?2000^2001 or 2001^2000 which is greater???????
[Up] View Post
hsbhatt (5015)

Forum Expert Blazing goIITian

Olaaa!! Perrrfect answer. 945  [1090 rates]
hsbhatt's Avatar
total posts: 1509    
online Online

One way is to consider the function f(x) = x^{\frac{1}{x}}. Prove that for x>e, this is a decreasing function.


It is also a useful fact to remember the order in the following series


1<\sqrt [2] 2< \sqrt [3] 3>\sqrt [4] 4>\sqrt [5] 4>...>\sqrt [n] n>....


Now, you can use this result to finish the problem
Time wounds all heels
 this reply: 22 points  (with 4 Olaaa!! Perrrfect answer.   in 5 votes )   [?]
 
You have to be logged on to rate
  
 
Forum Index -> Algebra
Go to:   

Top Offers for goIITians
Correspondence Courses
Brilliant Tutorials
Narayana Institute
Aakash Institute
Classroom/Crash Courses
Narayana - Kota , Delhi , Others
Brilliant Tutorials - Class , Crash
Aakash Institute - Medical , Engg
Online Test Series
Brilliant Tutorials
Narayana Institute
Aakash Institute
Mahesh Tutorials
AMITY      Sri Chaitanya