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 Post subject: exam problem.Posted: Mon, 6 Feb 2012 19:26:20 UTC
 Senior Member

Joined: Sun, 25 Dec 2011 00:28:25 UTC
Posts: 95
Hello.

I've posted pretty much a similar question time ago, and according to that, here's what I've done.

Attempted:

I need to know if I've done it right, Thank you.

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 Post subject: Re: exam problem.Posted: Mon, 6 Feb 2012 19:33:16 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12098
Location: Austin, TX
This problem does not have a well-defined answer.

For example, I could choose and and get another set of numbers which have a different sum.

However, it seems they wanted you to assume something like "a and b are both integers", and in that case you are safe.

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 Post subject: Re: exam problem.Posted: Mon, 6 Feb 2012 20:07:36 UTC
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Joined: Mon, 19 May 2003 19:55:19 UTC
Posts: 7949
Location: Lexington, MA
Hello, kreshnik!

You don't need logs for this probem . . .

Quote:

. .

We have: .

. . Hence: .

Got it?

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 Post subject: Re: exam problem.Posted: Mon, 6 Feb 2012 22:14:56 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12098
Location: Austin, TX
Soroban wrote:
Hello, kreshnik!

You don't need logs for this probem . . .

Quote:

. .

We have: .

. . Hence: .

Got it?

It's still not a unique choice.

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 Post subject: Re: exam problem.Posted: Tue, 7 Feb 2012 01:16:13 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Fri, 1 Jul 2011 01:17:26 UTC
Posts: 321
Soroban wrote:
Hello, kreshnik!

You don't need logs for this probem . . .

Quote:

. .

We have: .

. . Hence: .

Got it?

It's still not a unique choice.

Why not?

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 Post subject: Re: exam problem.Posted: Tue, 7 Feb 2012 01:21:30 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12098
Location: Austin, TX
mathematic wrote:
Soroban wrote:
Hello, kreshnik!

You don't need logs for this probem . . .

Quote:

. .

We have: .

. . Hence: .

Got it?

It's still not a unique choice.

Why not?

See my very first post in this topic.

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 Post subject: Re: exam problem.Posted: Tue, 7 Feb 2012 06:05:57 UTC
 Member

Joined: Mon, 6 Feb 2012 05:44:39 UTC
Posts: 27
I think you have to answer this with some insight on the context of this question - I don't see where that was given in the OP.

is is from a basic course dealing with rational numbers where numbers are represented uniquely with the prime power factorization?

or is it from a more advanced course that would expect the use of logs to solve it?

we have 2 variables a and b and only one equation to work with - restricting the answer to be "integers" (ie prime power factorization) there is only going to be one answer but over the reals there are an infinite number of solutions

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 Post subject: Re: exam problem.Posted: Tue, 7 Feb 2012 09:10:16 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12098
Location: Austin, TX
jules wrote:
I think you have to answer this with some insight on the context of this question - I don't see where that was given in the OP.

is is from a basic course dealing with rational numbers where numbers are represented uniquely with the prime power factorization?

or is it from a more advanced course that would expect the use of logs to solve it?

we have 2 variables a and b and only one equation to work with - restricting the answer to be "integers" (ie prime power factorization) there is only going to be one answer but over the reals there are an infinite number of solutions

It's my policy not to make assumptions other than those from what forum the topic is posted in.

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 Post subject: Re: exam problem.Posted: Tue, 7 Feb 2012 20:20:50 UTC
 Senior Member

Joined: Sun, 25 Dec 2011 00:28:25 UTC
Posts: 95
Hello.

Thank you very much guys, don't worry shadow we're not dealing with such difficult problems yet!
I think Soroban just did an easier way of doing it (comparing with mine ).

By the way:
Code:
\cdot

Is cooler!

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 Post subject: Re: exam problem.Posted: Wed, 8 Feb 2012 01:01:07 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Fri, 1 Jul 2011 01:17:26 UTC
Posts: 321
mathematic wrote:
Soroban wrote:
Hello, kreshnik!

You don't need logs for this probem . . .

Quote:

. .

We have: .

. . Hence: .

Got it?

It's still not a unique choice.

Why not?

See my very first post in this topic.

I see your point. I was assuming a and b had to be integers.

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