- #1

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**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

- Thread starter vadlamudit
- Start date

- #1

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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

Paradox

Sorry, you realised it before I did.

- #3

Paradox

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

ab = a

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

dividing by a;

b-1 = 1

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

- #4

dg

Good until here....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}

Not good the first passage... it would be

dividing by a;

b-1 = 1^{b-2}= 1 => b = 2

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

b-1 = a

- #5

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

So, a

But, b*b(a-1)

Now, 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

2

- #6

suffian

Agreed.Originally posted by Lonewolf

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

Not quite sure I follow here.So, a^{b}= b(a-1)^{b}

If a = b(a-1), then it follows that a

- #7

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Quite right. Well spotted.

- #8

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a=b(a-1)

a

ab=b

So, (a-1) = b

Then, 1 = b

This is true if b

b

- #9

suffian

I don't disagree, but that doesn't reallyOriginally 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.

edit:

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

a

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- #10

Thoth

From (1):a

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

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

- #11

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I think it's good up until here. The b should be in the brackets to makeln(b/(b-1))^{b}=ln(b2/(b-1))

ln([b/(b-1)]

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.

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- #12

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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?

- #13

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:(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)

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.

- #14

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(b^{2}/(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(b^{2}/(b-1) means this: b^{b}/(b-1)^{b}=b^{2}/(b-1). (By using e^{ln} definition). If b^{b}=b^{2} then b=2 and if (b-1)^{b}=(b-1)^{1} then b=1. however, we knew already that b cannot be 1.

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

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- #15

suffian

Unfortunantly, for that last equation to be true, bOriginally posted by Thoth

In ln(b^{b}/(b-1)^{b}))=ln(b^{2}/(b-1) means this: b^{b}/(b-1)^{b}=b^{2}/(b-1). (By using e^{ln}definition). If b^{b}=b^{2}then b=2 and if (b-1)^{b}=(b-1)^{1}then b=1. however, we knew already that b cannot be 1.

- #16

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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/a

b=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

a=b= 2 or 0

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

- #17

ahrkron

Staff Emeritus

Gold Member

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It is not!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)

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

suffian

- #19

suffian

from "a+b=ab"

0 + b = 0*b -> b=0 -> a

a + 0 = a*0 -> a=0 -> a

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

Continuing:

a

a

( b/(b-1) )

b

|b

|b|

ln|b|

(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)/( x

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.

Code:

```
+ + - +
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
```

Code:

```
+ - + + -
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
```

We can label these zeros as b

b

If we assume b

0<b

But this would mean that a

Anyways, now we have eliminated all but one possible solution, b

We now try setting b=2 and seeing if this really is a solution:

a = 2/(2-1) = 2

2 +2 = 2

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, b

(a=-3.14104155643, b=.75851486)

- #20

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

a

by the fundamental theorem of arithmetic, a = b

On one hand, y=a

On the other hand, y = b

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

Therefore a=b=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.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.

- #21

suffian

ok.Originally posted by KL Kam

If both a and b are integers

I understand that y (where y=a

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

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 = 2

And we have a integral solution in which both a and b are

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'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.

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

- #22

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a+b=ab=a^bAlso, 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

odd + odd = even

odd * odd = odd

so a, b can't be both odd

- #23

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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

- #24

Mentallic

Homework Helper

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It needs to also satisfy a^b.

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

- #25

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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

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