Tags: babab, probe
 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 [:((] [:((] [:((] [:((] [:((]
 P: n/a Sorry, you realised it before I did.
 P: n/a Okey doke, the (hopefully) non-erroneous method. a+b = ab => b = ab-a = a(b-1) ab = ab => b = ab-1 so, b = a(b-1) = ab-1 dividing by a; b-1 = 1b-2 = 1 => b = 2 a+b = ab => a+2 = 2a => a=b=2
P: n/a

 Originally posted by Paradox Okey doke, the (hopefully) non-erroneous method. a+b = ab => b = ab-a = a(b-1) ab = ab => b = ab-1 so, b = a(b-1) = ab-1
Good until here....
 dividing by a; b-1 = 1b-2 = 1 => b = 2 a+b = ab => a+2 = 2a => a=b=2
Not good the first passage... it would be
b-1 = ab-2
 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 = ab-b = b(a-1) So, ab = b(a-1)b But, b*b(a-1)b = ab, so, a = b*(a-1)b Now, b*(a-1)b = b(a-1), so (a-1)b = (a-1) 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 ab= ab = a+b 22 = 2*2 = 2+2, as required. Thus, a=b=2.
P: n/a
 Originally posted by Lonewolf Using a+b = ab, we get a = ab-b = b(a-1)
Agreed.

 So, ab = b(a-1)b
Not quite sure I follow here.
If a = b(a-1), then it follows that ab = ( b(a-1) )b = bb(a-1)b
 P: 333 Quite right. Well spotted.
 P: 333 I figured where I went wrong. It was my dodgy handwriting... a=b(a-1) ab=bb(a-1)b ab=b2(a-1)=bb(a-1)b So, (a-1) = bb-2(a-1)b Then, 1 = bb-2(a-1)b-1 This is true if bb-2 is 1, and (a-1)b-1 is 1. bb-2 = 1 if b-2=0, so b=2. (a-1)=1, so a=2.
P: n/a
 Originally posted by Lonewolf Then, 1 = bb-2(a-1)b-1 This is true if bb-2 is 1, and (a-1)b-1 is 1. bb-2 = 1 if b-2=0, so b=2. (a-1)=1, so a=2.
I don't disagree, but that doesn't really prove that there are no other numbers a and b for which bb-2 and (a-1)b-1 are inverses.

edit:

I think the problem essentially boils down to prove no real solutions other than a=2 for:

aa-2=(a-1)a-1
 P: n/a How about this? From (1):ab=ab we get (b)*ln(a)=ln(a)+ln(b) and From (2):a+b=ab we get a=b/(b-1) so ln(a)=ln(b/(b-1)) subsitute (2) into (1) to get: (b)*ln[(b/b-1)]=ln(b/(b-1))+ln(b) so ln(b/(b-1))b=ln(b2/(b-1)) which means that bb=b2 therefore, b=2 and from a=b/(b-1) we get a=2/(2-1)=2 [g)]
P: 333
 ln(b/(b-1))b=ln(b2/(b-1))
I think it's good up until here. The b should be in the brackets to make

ln([b/(b-1)]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.
 P: 513 I. a+b=ab I. b=ab-a I. b=a(b-1) I. a=b/(b-1) II. ab=a^b I into II. b^2/(b-1)=(b/(b-1))^b 2ln b - ln(b-1) = b ln b - b ln(b-1) (2-b)ln b = (1-b)ln(b-1) (2-b)/(1-b) = ln(b-1)/ln(b) Use definition of ln... (2-b)/(1-b) = lim(n->[oo]) n((b-1)^(1/n)-1) / n(b^(1/n)-1) n cancels ... (2-b)/(1-b) = lim(n->[oo]) ((b-1)^(1/n)-1) / (b^(1/n)-1) Using Bernoulli... (2-b)/(1-b) = lim(n->[oo]) (1 + b/n -1/n -1) / (1 + b/n -1) (2-b)/(1-b) = lim(n->[oo]) (b-1)/b (2-b)/(1-b) = (b-1)/b 2b-b^2 = 2b-b^2-1 0=-1 No solution. Wow. Where's the flaw? I know the RHS must be zero. Hopital, maybe?
P: n/a
 (2-b)/(1-b) = lim(n->inf) ((b-1)^(1/n)-1) / (b^(1/n)-1) Using Bernoulli... (2-b)/(1-b) = lim(n->inf) (1 + b/n -1/n -1) / (1 + b/n -1)
I think your flaw may be in making the transition above, I don't know Bernoulli's rule or how you changed the ln's into a limit, but the limits don't match up:

lim(n->inf) [ ((b-1)^(1/n)-1) / (b^(1/n)-1) ] = ln(b-1)/ln(b)

and

lim(n->inf) [ (1 + b/n -1/n -1) / (1 + b/n -1) ] = (b-1)/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.
 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/(b-1))b and what you wrote comes out the same, which is: ln(bb/(b-1)b))=ln(b2/(b-1). 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/(b-1) and ln(a)=ln(b/(b-1)) and ask what values for b make sense for a? Here a=b/(b-1) 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(bb/(b-1)b))=ln(b2/(b-1) means this: bb/(b-1)b=b2/(b-1). (By using eln definition). If bb=b2 then b=2 and if (b-1)b=(b-1)1 then b=1. however, we knew already that b cannot be 1. [:D]
P: n/a
 Originally posted by Thoth In ln(bb/(b-1)b))=ln(b2/(b-1) means this: bb/(b-1)b=b2/(b-1). (By using eln definition). If bb=b2 then b=2 and if (b-1)b=(b-1)1 then b=1. however, we knew already that b cannot be 1. [:D]
Unfortunantly, for that last equation to be true, bb does not need to equal b2 and (b-1)b need not equal (b-1). As long as the ratio between the pairs is satisfied, the equation can be satisfied and so we haven't yet proved no other real numbers can satisfy the last equation.
 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=ab-1 b-1 = b/a b=ab/a ba=ab 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 = a2 a=b= 2 or 0 I think the proof would be more elegant if we can do it using pure algebra.
Emeritus
PF Gold
P: 734
 Originally posted by KL Kam ... a log b = b log a log a / a = log b / b a=b (as the function f(x)=logx/x is strictly increasing)
It is not!

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.
 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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