1. RayorDragonFall

    OP RayorDragonFall I see you've played knifey-spoony before.
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    Ok so I have this:

    [​IMG]

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

    Thank you guys sooooo much [​IMG]

    PS. Fack I typoed the topic name XD
    Thats what happens when you have someone trying to get attention around you all the time [​IMG].
     
  2. Linkiboy

    Linkiboy GBAtemp Testing Area
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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

    Let me upload work....

    [​IMG]

    good luck reading it
     
  3. RayorDragonFall

    OP RayorDragonFall I see you've played knifey-spoony before.
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    Eeny meeny miny mo says its A, but I'm not totally sure lol XD
     
  4. Zaago

    Zaago Member
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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.

    I hope that can help you.


    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. RayorDragonFall

    OP RayorDragonFall I see you've played knifey-spoony before.
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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

    dawn.wan GBAtemp Fan
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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

    Linkiboy GBAtemp Testing Area
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    Nope.

    N is already an exponent, so you just multiply.
     
  8. shakirmoledina

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

    deathfisaro Narcistic Deathfisaro Fan
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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

    Prophet Resident Black Militant
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    Stupid math, always being hard and stupid... *goes and writes a sonnet*
     
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