Please Help! given a+b=ab=a^b, probe a=b=2

vadlamudit

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 [:((] [:((] [:((] [:((] [:((]

Paradox

Sorry, you realised it before I did.

Paradox

Okey doke, the (hopefully) non-erroneous method.

a+b = ab => b = ab-a = a(b-1)

ab = a^b => b = a^b-1

so, b = a(b-1) = a^b-1

dividing by a;

b-1 = 1^b-2 = 1 => b = 2

a+b = ab => a+2 = 2a => a=b=2

dg

Good until here....
Not good the first passage... it would be
b-1 = a^b-2

Lonewolf

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, a^b = 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 a^b= ab = a+b
2² = 2*2 = 2+2, as required. Thus, a=b=2.

suffian

Agreed.

Not quite sure I follow here.
If a = b(a-1), then it follows that a^b = ( b(a-1) )^b = b^b(a-1)^b

Lonewolf

Quite right. Well spotted.

Lonewolf

I figured where I went wrong. It was my dodgy handwriting...

a=b(a-1)
a^b=b^b(a-1)^b
ab=b²(a-1)=b^b(a-1)^b
So, (a-1) = b^b-2(a-1)^b
Then, 1 = b^b-2(a-1)^b-1
This is true if b^b-2 is 1, and (a-1)^b-1 is 1.
b^b-2 = 1 if b-2=0, so b=2. (a-1)=1, so a=2.

suffian

I don't disagree, but that doesn't really prove that there are no other numbers a and b for which b^b-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:

a^a-2=(a-1)^a-1

Thoth

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/(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

b^b=b² therefore,

b=2 and from a=b/(b-1) we get a=2/(2-1)=2
[g)]

Lonewolf

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.

arcnets

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?

suffian

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.

Thoth

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(b^b/(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(b^b/(b-1)^b))=ln(b²/(b-1) means this: b^b/(b-1)^b=b²/(b-1). (By using e^ln definition). If b^b=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]

suffian

Unfortunantly, for that last equation to be true, b^b 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.

KLscilevothma

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^b-1
b-1 = 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 = a²
a=b= 2 or 0

I think the proof would be more elegant if we can do it using pure algebra.

ahrkron

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.