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**tony123****Member**- Registered: 2007-08-03
- Posts: 189

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Why did the vector cross the road?

It wanted to be normal.

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**tony123****Member**- Registered: 2007-08-03
- Posts: 189

Tank you mathsyperson but

let x=6/10

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Sorry, I'm not sure what you're getting at.

6/10 is a solution and is in my solution set, so there's no contradiction.

If you're hinting at a more elegant method of reaching the answer, I'm afraid I don't get that either.

Why did the vector cross the road?

It wanted to be normal.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Perhaps hes hinting that your solution isnt wholly correct. *x* = 6⁄10 is **NOT** a solution. Also, you cant have *x* = 1. What is logarithm to base 1, anyway?

Anyway, log[sub]*x*[/sub] where *x* < 1 problematic. If *x* < 1, then you actually have log[sub]*x*[/sub]*y* > 0 when *y* < 1 and log[sub]*x*[/sub]*y* < 0 when *y* > 1. With this in mind, I think youll find that not only

is invalid but also

is a solution set as well.

*Last edited by JaneFairfax (2007-12-23 01:00:39)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Ah, fair enough. I made too many assumptions in the first step.

Why did the vector cross the road?

It wanted to be normal.

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