# Solve a short maths problem and win an internet cookie!

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1. ### OP RayorDragonFall I see you've played knifey-spoony before. Member Level 3

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Ok so I have this:

So pretty much, x^-n = 1/y^2 is a, b, c or d.

Thank you guys sooooo much

PS. Fack I typoed the topic name XD
Thats what happens when you have someone trying to get attention around you all the time .

2. ### Linkiboy GBAtemp Testing Area Member Level 6

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Schools over for me and I already forgot all of this, but I think its ... nevermind I forgot how to do these

EDIT: i did it, got a

3. ### OP RayorDragonFall I see you've played knifey-spoony before. Member Level 3

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Eeny meeny miny mo says its A, but I'm not totally sure lol XD

4. ### Zaago Member Newcomer Level 2

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Has being so long since school but if I remember correctly:

Log(base) n = e :: base^e = n
so then the most acurrated answer is C because

Log(x) y = -2n :: x^(-2n) = y

How to pass the exponent (2) to the other side?, I don't remember but that seems the most logic answer to me.

EDIT: Wait, now that I think about it the answer also could be A because:

Log(x) y = n/2 :: x^(n/2) = y

Where is the (-)?, I have no idea but now you have two posibles answers out of four.

5. ### OP RayorDragonFall I see you've played knifey-spoony before. Member Level 3

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Linki, I got it but I don't understand how raising to the -1 power at the end would make n positive? Wouldn't it be 1/-n?

6. ### dawn.wan GBAtemp Fan Member Level 1

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A is right, you simplify -1/n by factoring out -1.. it's just one of those fundamental principles about roots cannot be negative..

7. ### Linkiboy GBAtemp Testing Area Member Level 6

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Nope.

N is already an exponent, so you just multiply.

8. ### shakirmoledina Legend Member Level 6

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A.. pretty easy
Log y X + 2Log x Y = 4
XY=9
Note : the small y and x mean that they are the base

9. ### deathfisaro Narcistic Deathfisaro Fan Member Level 3

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Here are two rules:
log(ab) = log(a) + log(b)
log(a^2) = 2log(a) (can be derived from the formula above, use another a in b's place)

x^(-n) = 1 / (y^2)
This can be re-written as
x^(-n) * y^2 = 1

1) Take log base x on both sides
Use the first rule I wrote above, and log(1) = 0 for any base of log.

logx (x^(-n)) + logx (y^2) = 0

logx (x^(-n)) is just -n, logx (y^2) can be re-written as 2 logx(y) (Look at the second rule above)
So,
-n + 2 logx(y) = 0

add n to both sides and divide both sides by 2
logx(y) = n / 2

2) Take log base y on both sides
logy(x^(-n)) + logy(y^2) = 0

-n logy(x) + 2 = 0

subtract 2 from both sides and divide both sides by -n

logy(x) = 2/n

Edit: Wait, regardless of which one you do first, you have to do log base x anyway. =P

10. ### Prophet Resident Black Militant Member Level 3

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Stupid math, always being hard and stupid... *goes and writes a sonnet*

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