Can you help me prove a=b=2 for the equation a+b=ab=a^b?

[vadlamudit]

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

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 realized 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]

Originally posted by 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

Good until here...

dividing by a;

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

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

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]

Originally posted by Lonewolf

Using a+b = ab, we get a = ab-b = b(a-1)

Agreed.

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

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]

Originally posted by Lonewolf
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.

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

 

[Lonewolf]

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

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]

(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.

 

[Thoth]

Lonewolf, you are probably right and I might need to change my classes . But as far as I can tell , 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.

 

[suffian]

Originally posted by Thoth

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.

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]

My teacher gave me an elegant solution

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]

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.

 

[suffian]

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.

 

[suffian]

First note that neither a nor b can be equal to zero or one:

from "a+b=ab"
0 + b = 0*b -> b=0 -> a^b = 0⁰ (an undefined quantity)
a + 0 = a*0 -> a=0 -> a^b = 0⁰
1 + b = 1*b -> contradiction
a + 1 = a*1 -> contradiction

By establishing that a,b != {0,1}, we are now free to divide by {a, b, a-1, b-1}

a*b = a + b
a*b - a = b
a*(b-1) = b
a = b/(b-1)

Let's pause for a moment and further realize that b cannot be less than zero:

if (b<0) -> [a = -|b|/(-|b|-1)] -> (a>0)
but this means a*b is negative and a^b is postive, which is a contradiction since they must equal each other.

Continuing:
a^b = a*b
a^b-1 = b
( b/(b-1) )^b-1 = b (substitution from above)
b^b-2 = (b-1)^b-1
|b^b-2| = |(b-1)^b-1|
|b|^b-2 = |b-1|^b-1
ln|b|^b-2 = ln|b-1|^b-1
(b-2)ln|b| - (b-1)ln|b-1| = 0

We now set f(x) = (x-2)ln|x| - (x-2)ln|x-1|, x!={0,1}. The zeros of f(x) are the only possible solutions for b as f(b) must equal zero.

Finding these zeros is a difficult task and requires analyzing the derivatives.

f'(x) = ln|x/(x-1)| - 2/x, x!={0,1}
f''(x) = (x-2)/( x²(x-1) ), x!={0,1}

Since f''(x) is the only function we really recognize, we us it to find out what f'(x) looks like and then we use that to find out what f(x) looks like.

            +        +          -        +
f''(x) <---------|---------|---------|--------->
                 0(und)    1(und)    2
Lim[ x-> +/-infinity, f'(x) ] = 0
f'(0.1) < 0 and f'(0.95) > 0 (just plug it in)
f'(1.1) > 0 and f'(1.90) < 0

This means that f'(x) must have only two zero which exist on (0,1) and (1,2) (because f' is continuous within the intervals inbetween) which we will label c₁ and c₂ respectively. This in turn implies that the sign chart of f'(x) looks as follows:

           +         -         +         +         -    
f'(x) <---------|---------|---------|---------|--------->
                0(und)    c1        1(und)    c2(c2<2)
Limit[ x->1, f(x) ] = 0 (though not defined at 1)
f(-4) < 0 and f(-3) > 0
f(0.1) > 0 and f(0.9) < 0
f(1.1) > 0 and f(2.1) < 0

Finally, we use the last set of information to note that there are precisely three zeros of f(x). Since limit of f(x) as x approaches 1 is zero and f(x) is increasing on c₁ to c₂ there can't be a zero within this interval. However the other three intervals do have zeros because f(x) is continuous on them and changes sign: (-inf, 0), (0, c₁), (c₂, +inf).
We can label these zeros as b₁, b₂, and b₃ respectively.

b₁, b₂, and b₃ are the only possible solutions for b because f(b) must equal zero as I stated above. But also remember that b > 0 so that b₁, which is in (-inf, 0), is ruled out.

If we assume b₂ is correct than a is negative:

0<b₂<c₁<1 -> a = |b|/(|b|-1) = pos/neg = neg.

But this would mean that a^b has a negative base. This leads to a complex number for a^b which certainly cannot equal a+b. (Although if you simply take the neg base as meaning -|a|^b, then this actually would look like a solution as i tried plugging in numbers and it comes extremely close).

Anyways, now we have eliminated all but one possible solution, b₃, which exists in the interval [c₂, +inf) where c₂ is less than 2. By inspection we know f(2) = 0 and therefore b₃ must equal 2.

We now try setting b=2 and seeing if this really is a solution:
a = 2/(2-1) = 2
2 +2 = 2² = 2*2 = 4

I know this has to be the ugliest solution you would ever want to see, and I've omitted some of the stuff to keep this from being any longer... but i think this really proves that (a=2,b=2) is the only solution.

Actually, like i said, b₂ might work. Switch your calculator out of complex mode and into real mode and try it out:
(a=-3.14104155643, b=.75851486)

 

[KLscilevothma]

a, b integers ?

If both a and b are integers, then a must equal to b.

a^b=b^a
by the fundamental theorem of arithmetic, a = b

On one hand, y=a^b where y is an integer and it can be uniquely factorized. (a is a prime number)

On the other hand, y = b^a where b is a prime number.

We can prove my contradition that a and b are even numbers.

Therefore a=b=2

quote:
--------------------------------------------------------------------------------
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.

I'm sorry to cause confusion here. Yes, f(x) is strictly increasing when e>x>0 and it is strictly decreasing when x>e. It seems that we have infinitely many solutions if a and b aren't integers.

 

[suffian]

Originally posted by KL Kam
If both a and b are integers

ok.

On one hand, y=a^b where y is an integer and it can be uniquely factorized. (a is a prime number)

On the other hand, y = b^a where b is a prime number.

We can prove my contradition that a and b are even numbers.

Therefore a=b=2

I understand that y (where y=a^b) can be uniquely factorized, but I don't see how that means a^b is the unique factorization of y. For example 729 = 9³ but its unique factorization is 3⁶. In other words, I don't see why a and b have to be prime numbers.

Also, how do you prove that a, b must be even? I understand that a,b must both be even or odd for both sides to be even or odd respectively but not why they must just be even.

Also, if we take a=2 and b=4, then
16 = 2⁴ = a^b = b^a = 4² = 16
And we have a integral solution in which both a and b are not prime and not both 2.

I'm sorry to cause confusion here. Yes, f(x) is strictly increasing when e>x>0 and it is strictly decreasing when x>e. It seems that we have infinitely many solutions if a and b aren't integers.

To the problem ln(a)/a = ln(b)/b, yeah, i agree but this doesn't mean the original problem has infinitely many solutions (see my post right above your last one).

I hope I don't sound like I'm rebuking you, I certainly don't mean to.

 

[KLscilevothma]

Also, how do you prove that a, b must be even? I understand that a,b must both be even or odd for both sides to be even or odd respectively but not why they must just be even

a+b=ab=a^b

odd + odd = even
odd * odd = odd
so a, b can't be both odd

 

[budijoe]

I have this simple solution.

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

So, if you just input b value, you will know a value?

Is this the question? Or i am just stupid? LOL

 

[Mentallic]

It needs to also satisfy a^b.

Damn a 7 year old thread? I didn't even know this forum was around in '03.

 

[sailaja.g]

It is easy

given a+b=ab=a^b

ans:
ab=a^b
ba=b^a
=> a^b=b^a
=> a=b ...(1)

a+b=ab
from (1)
=>b+b=b*b
=>2b=b^2
b cancel both sides
=>b=2
from (1)
a=b=2

I think I am right

 

[Mentallic]

It is easy

given a+b=ab=a^b

ans:
ab=a^b
ba=b^a

No this is wrong.

$$3\cdot 1 = 3^1$$

but

$$1\cdot 3 \neq 1^3$$

 

[sailaja.g]

I would like to remember one fact
x^2-3*x+2=0 then x=1,2
x^2-3*x+2=0 if u take other than 1 & 2 this equation don't satisfy
ex x=3

3^2-3*3+2=0
2=0

it doesn't satisfy other than solutions for that equation

those are just eqs just like our above example
a^b=ab=ba=b^a
here I didn't assume anything I came for fact i.e, ab=ba

they gave that is true. our duty is to find values which satisfy those conditions
here our solution is 2. other than 2 no number can satisfy those conditions

Nice doubt thanks for make answer for that nice question