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**PCHydrogen****Member**- Registered: 2016-03-09
- Posts: 2

**I'm looking for help with the following question and the possible values for x and y, if any. If you do reply, please show your working out as I'd like to know how you've come to your answer. I've attempted it several times with a form of logarithms and algebra but every time I don't get a result.**

**If you do manage to get an answer, please inform me how!**

**Thank you!**

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 115

This system has in fact a solution and it can be solved as illustrated below:

From the first and the second equations we have:

And

So it follows that

Hence we have

Therefore x=0 and y=∞.

Q.E.F

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,039

Grantingriver wrote:

Therefore x=0 and y=∞.Q.E.F

This is obviously wrong, but I cannot decide if purposefully so...

*Last edited by anonimnystefy (2016-03-09 12:41:57)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 115

I hate to disagree with you, but in my opinion it is correct!! Let's solve the problem in different way: suppose we have an educated guess that the solutions are x=0 and y=∞ so to check that our supposed solutions are correct, we just have to see if they satisfy the system of equations or not. Let's do that:

Therefore the porposed solutions are correct.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 115

Dear anonimnystefy, we can not substitute directly in the given system of equations because one to the power of infinity is "indeterminate" http://www.vitutor.com/calculus/limits/ … inity.html so we have to modify the equation first in order to reach the correct conclusion. Thank you for your passing.

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,039

That's not really how equations work.

What Nehushtan wrote in post #2 is the correct solution.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 115

If someone present a solution to a problem, you just have to ask him for a proof of his results. if he gave you a correct argument, but you did not accept it, then you have to provide plausible reasons to justify your situation. And I think you did not do that. "Any one can reject any angument even if it is correct without reason". Sorry but your situation is weak.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,452

When I first looked at this (before Nehushtan's post) I thought it had no solutions. So I graphed the equations. The two graphs do indeed tend towards a crossing as x approaches 0 and y approaches infinity. But I wouldn't give that as a 'solution' because infinity cannot safely be regarded as a number. All sorts of problems arise if you write x equals infinity.

So I'm with Nehushtan. No solutions, or no finite solutions if you prefer.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,039

I try to stay away from argumentation of any sort since I've found they are more often fruitless than not.

To say that an ordered pair (x,y) satisfies the equations given in post #1 means that, when we substitute the values, we get a true equality. If you try to reduce the equations the way you did it, you cannot guarantee that the solutions are the same. It'd be like saying that the equation

where represents the function that maps the number x to its principal square root, has the solution 1 because squaring it you get the equation x=1, which the number 1 does satisfy.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 115

Bob bundy has settled the problem: If we seek a finite solutions to the given system of equations then there are "no" solutions, but if we want solutions without restrictions (which is in fact the case of this problem) then the previous arguments provide "correct" solutions (also this can be proven graphically). Finally if I give you the following equation:

And ask you: Is there a value which x can take and satisfies this equation? If you said "yes" then we have done (all the previous arguments are correct). And if you said "No" we have also done, since why we have to discuss the problem farther while you do not believe in the existence of infinity?!!

My respect to you and all who contributed in this topic.

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**frelo****Member**- Registered: 2016-03-12
- Posts: 2

Would you say that the same solution is also a solution of this system? Because all your equations hold also with this system, you just have to replace you ln 2 by ln 3 and ln 0.5 by ln 1/3.

This would be very cool, because this would imply 2=3.

(In other words, I agree with anonimnystefy)

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 115

Sorry but it does not imply that 2=3!! To show you why this is the case, consider the following two systems:

2x+3y=2

4x-y=3

And

2x+3y=3

4x-y=4

Both of these systems have solutions and they are different. However, this does not imply that 2=3 and 3=4 because 2 and 3 equals 2x+3y and 3 and 4 equals 4x-y respectively. On the other hand, the one who said that the solution of a system of equations are the values which make each equation in the system hold is the definition of the solutions of a system of equations: "the solutions of a system of equations are the values which satisfy each equation simultaneously". Finally who did say that the solutions of a system of equations, if they exist, have to satisfy one and only one system. This claim has to be poven, in fact, the provided system is a counter-example which disprove that claim.

Note: If you would like to exclude the deduced solutions form being a solution of the provided system, you have to modify the definition of the solutions of a system of equations to include the uniqueness to that system which are not included in the actual definition. Also in that case, we would have a solution to the system according to the previous and unmodified definition. "It's a matter of definition sir".

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**frelo****Member**- Registered: 2016-03-12
- Posts: 2

I completly agree with your exemple, however, I don't think you understood my argument. The big diference is that in the exemple you give, x and y have diferent value in both system, wether in my case, they have the same solution in both system. The solution of the system I proposed is, according to your argument, x=0 and y = ∞.

So both system have the same solution.

Of course, it's in general not a problem that two systems have the same solution. What is a problem is that, substituing in the same funcion the same number you sometime get 2 and sometime get 3.

In other word, if you replace in (1+x)^y x by 0 and y by ∞, do you get 2 or 3?

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**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 115

I didn't discuss this point because I already have talked about it (see #6). However, let's discuss it again in details. If we substitute in the given system of equations we would have:

And

Also we would get the same results for the system of equations which you proposed that is

and . The same situation occurs if we replace the numbers on the RHS of the first and the second equations of the original system with and respectively, where "a" is any real, or more generally, complex number hence these values (x=0 and y=∞) are solutions of a family of infinite number of systems of equations. This stem from the fact that does "not" equal "1" it is "indeterminate" that is it could have any values depending on the problem considered, therefore it satisfies any system belong to this family. To prove that you can take the logarithm of both sides for each equation and follow the procedures which are illustrated above.Note: For the sake of cosistency and copmleteness, mathematicians have cosidered the following quantities as indeterminates:

I think this and the pevious discussions answer all of your questions.

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