
#1
Apr2403, 08:27 AM

P: 12

Please help me on this:
Given a+b=ab=a^b prove a=b=2 it seems like a very simple task but I have lost my sleep for the past couple of nights over it. please help or ill be [:((] [:((] [:((] [:((] [:((] 


#2
Apr2403, 06:52 PM

P: n/a

Sorry, you realised it before I did.



#3
Apr2403, 07:11 PM

P: n/a

Okey doke, the (hopefully) nonerroneous method.
a+b = ab => b = aba = a(b1) ab = a^{b} => b = a^{b1} so, b = a(b1) = a^{b1} dividing by a; b1 = 1^{b2} = 1 => b = 2 a+b = ab => a+2 = 2a => a=b=2 


#4
Apr2403, 07:25 PM

P: n/a

Please Help! given a+b=ab=a^b, probe a=b=2b1 = a^{b2} 



#5
May2403, 12:38 PM

P: 333

This may well be a bit late, but I was bored and browsing through some old posts.
Using a+b = ab, we get a = abb = b(a1) So, a^{b} = b(a1)^{b} But, b*b(a1)^{b} = ab, so, a = b*(a1)^{b} Now, b*(a1)^{b} = b(a1), so (a1)^{b} = (a1) This is true if a=2, or b=1. Substituting b=1 into ab = a+b, a*1 = a+1, which leads to 0=1, a contradiction, thus b is not 1. Substituting a=2 into ab = a+b, 2b = b+2, so b=2. Substituting these two values into a^{b}= ab = a+b 2^{2} = 2*2 = 2+2, as required. Thus, a=b=2. 


#6
May2403, 04:34 PM

P: n/a

If a = b(a1), then it follows that a^{b} = ( b(a1) )^{b} = b^{b}(a1)^{b} 



#7
May2403, 04:42 PM

P: 333

Quite right. Well spotted.




#8
May2403, 05:06 PM

P: 333

I figured where I went wrong. It was my dodgy handwriting...
a=b(a1) a^{b}=b^{b}(a1)^{b} ab=b^{2}(a1)=b^{b}(a1)^{b} So, (a1) = b^{b2}(a1)^{b} Then, 1 = b^{b2}(a1)^{b1} This is true if b^{b2} is 1, and (a1)^{b1} is 1. b^{b2} = 1 if b2=0, so b=2. (a1)=1, so a=2. 


#9
May2403, 09:31 PM

P: n/a

edit: I think the problem essentially boils down to prove no real solutions other than a=2 for: a^{a2}=(a1)^{a1} 


#10
May2503, 01:52 AM

P: n/a

How about this?
From (1):a^{b}=ab we get (b)*ln(a)=ln(a)+ln(b) and From (2):a+b=ab we get a=b/(b1) so ln(a)=ln(b/(b1)) subsitute (2) into (1) to get: (b)*ln[(b/b1)]=ln(b/(b1))+ln(b) so ln(b/(b1))^{b}=ln(b^{2}/(b1)) which means that b^{b}=b^{2} therefore, b=2 and from a=b/(b1) we get a=2/(21)=2 [g)] 



#11
May2503, 03:39 AM

P: 333

ln([b/(b1)]^{b}) This problem is harder than it first seems [s(] There are other inverses to my equation after all...take b=1, then a could be any real number, except 1...oh well. I think Thoth's onto something though. 



#12
May2503, 04:55 PM

P: 513

I. a+b=ab
I. b=aba I. b=a(b1) I. a=b/(b1) II. ab=a^b I into II. b^2/(b1)=(b/(b1))^b 2ln b  ln(b1) = b ln b  b ln(b1) (2b)ln b = (1b)ln(b1) (2b)/(1b) = ln(b1)/ln(b) Use definition of ln... (2b)/(1b) = lim(n>[oo]) n((b1)^(1/n)1) / n(b^(1/n)1) n cancels ... (2b)/(1b) = lim(n>[oo]) ((b1)^(1/n)1) / (b^(1/n)1) Using Bernoulli... (2b)/(1b) = lim(n>[oo]) (1 + b/n 1/n 1) / (1 + b/n 1) (2b)/(1b) = lim(n>[oo]) (b1)/b (2b)/(1b) = (b1)/b 2bb^2 = 2bb^21 0=1 No solution. Wow. Where's the flaw? I know the RHS must be zero. Hopital, maybe? 


#13
May2503, 05:33 PM

P: n/a

lim(n>inf) [ ((b1)^(1/n)1) / (b^(1/n)1) ] = ln(b1)/ln(b) and lim(n>inf) [ (1 + b/n 1/n 1) / (1 + b/n 1) ] = (b1)/b I would be a little hesitant to use ln's anyway because you don't know whether the quantities are negative. But it looks like a new avenue. 


#14
May2603, 01:21 AM

P: n/a

Lonewolf, you are probably right and I might need to change my classes[a)]. But as far as I can tell[8)], the result of both ln(b/(b1))^{b} and what you wrote comes out the same, which is: ln(b^{b}/(b1)^{b}))=ln(b^{2}/(b1). However I give your opinion the benefit of the doubt since typing mathematics does not come easy for me[:((].
A major factor is to look at a=b/(b1) and ln(a)=ln(b/(b1)) and ask what values for b make sense for a? Here a=b/(b1) first we notice that b cannot be =1 because of the singularity at b=1. After noticing this we then have to ask how about if b>1 and b<1 what would happen then. As b increases in value how a is being effected. We also need to know what values for both a and b satisfies the given conditions in the problem. In ln(b^{b}/(b1)^{b}))=ln(b^{2}/(b1) means this: b^{b}/(b1)^{b}=b^{2}/(b1). (By using e^{ln} definition). If b^{b}=b^{2} then b=2 and if (b1)^{b}=(b1)^{1} then b=1. however, we knew already that b cannot be 1. [:D] 


#15
May2603, 09:57 AM

P: n/a





#16
May2803, 02:37 AM

P: 321

This question has troubled me for some time, finally I gave it to my math teacher and he gave me an elegant solution today.
b=a^{b1} b1 = b/a b=a^{b/a} b^{a}=a^{b} a log b = b log a log a / a = log b / b a=b (as the function f(x)=logx/x is strictly increasing) =>2a = a^{2} a=b= 2 or 0 I think the proof would be more elegant if we can do it using pure algebra. 



#17
May2803, 09:30 AM

Emeritus
PF Gold
P: 734

f(1/e) = log(1/e)/e = log(e)/e = 1/e f(e) = log e / e = 1/e f(e^2) = log(e^2)/(e^2) = 2/e^2 =(2/e)*(1/e) < 1/e i.e., it goes up and then back down. 


#18
May2803, 01:11 PM

P: n/a

hey guys, i'm quite sure i solved the problem, but since i'm busy i'll try to post a formal proof by the end of the day.



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