I really don't know where else to turn and could use some Asian insight to a problem I have encountered...

  • Posted by a hidden member.
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    Jul 22, 2012 9:41 PM GMT
    Hello Asians,

    Thank you for clicking on my thread. Let me start off by saying I greatly appreciate the help for this problem I have encountered. I tried everything I could to solve it, but am now at the point where I need help, so I have come to you.

    Here is the problem:

    Which of the following functions f is f(x) = f(1-x) for all of x?

    1) f(x)=1-x
    2) f(x)=1-x^2
    3) f(x)=x^2-(1-x)^2
    4) f(x)=(x^2)(1-x)^2
    5) f(x)=x/1-x

    Like I have said, I tried everything to figure out this problem and realize that I need some special insight. If you could tell me the answer and, more importantly, how to find the answer I would love you a long time.

    Thanks again, and you guys are the best!
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    Jul 22, 2012 9:51 PM GMT
    Chainers saidHello Asians,

    Thank you for clicking on my thread. Let me start off by saying I greatly appreciate the help for this problem I have encountered. I tried everything I could to solve it, but am now at the point where I need help, so I have come to you.

    Here is the problem:

    Which of the following functions f is f(x) = f(1-x) for all of x?

    1) f(x)=1-x
    2) f(x)=1-x^2
    3) f(x)=x^2-(1-x)^2
    4) f(x)=(x^2)(1-x)^2
    5) f(x)=x/1-x

    Like I have said, I tried everything to figure out this problem and realize that I need some special insight. If you could tell me the answer and, more importantly, how to find the answer I would love you a long time.

    Thanks again, and you guys are the best!


    It's basically asking which function(s) out of the five options will be equal to f(1-x).

    So, first you have to understand what f(1-x) means. It's similar to finding f(2). You would usually plug in 2 for x, right? Well, try plugging in x=1-x in each of the five functions.

    So, in essence you want to replace each of the x variables in each of your five functions by (1-x) and solve the algebra and figure out which one ends up giving f(x) again.

    (and no I didn't just join to answer this lol. i just happened to return today and this just happens to be my first post)
  • Posted by a hidden member.
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    Jul 22, 2012 9:51 PM GMT
    ^^no Im saying the fucking GMAT is racist because this is math that only an Asian can do and it is fucking horse shit.

    Thats what I am saying.
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    Jul 22, 2012 9:55 PM GMT
    So do you get it?
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    Jul 22, 2012 9:56 PM GMT
    sirdreamalott saidSo do you get it?


    See thats what I thought it said and I did that but I didnt find one that ended up equaling itself.

    Maybe I just did it wrong but I didnt really find a correct answer to this problem.
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    Jul 22, 2012 9:58 PM GMT
    I've been trying to think of a name for my new punk band, and I'd like to thank the OP for it's new name......Asian Insight! We'll all be performing while wearing hospital masks of course.
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    Jul 22, 2012 10:01 PM GMT
    Chainers said
    sirdreamalott saidSo do you get it?


    See thats what I thought it said and I did that but I didnt find one that ended up equaling itself.

    Maybe I just did it wrong but I didnt really find a correct answer to this problem.


    The answer is D.

    Here's why:

    The function in D is: f(x) = x^2 (1-x)^2

    Plugging in x=(1-x) into D gives:

    f(x) = (1-x)^2 (1-1+x)^2
    = (1-x)^2 x^2
    which is exactly the same function in D. Voila!
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    Jul 22, 2012 10:02 PM GMT
    sirdreamalott said
    Chainers said
    sirdreamalott saidSo do you get it?


    See thats what I thought it said and I did that but I didnt find one that ended up equaling itself.

    Maybe I just did it wrong but I didnt really find a correct answer to this problem.


    The answer is D.

    Here's why:

    The function in D is: f(x) = x^2 (1-x)^2

    Plugging in x=(1-x) into D gives:

    f(x) = (1-x)^2 (1-1+x)^2
    = (1-x)^2 x^2
    which is exactly the same function in D. Voila!


    Thank you so much for the help.

    I have another one if you dont mind, its to find the value for M.

    (1/5)^m*(1/4)^17=1/2(10^32)

    How do you find M with this abomination?
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    Jul 22, 2012 10:03 PM GMT
    gif.latex?f(x)=x^2(1-x)^2%20---So,%20f(1

    Just a prettier latex version as to why it's 4.
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    Jul 22, 2012 10:06 PM GMT
    Chainers said
    sirdreamalott said
    Chainers said
    sirdreamalott saidSo do you get it?


    See thats what I thought it said and I did that but I didnt find one that ended up equaling itself.

    Maybe I just did it wrong but I didnt really find a correct answer to this problem.


    The answer is D.

    Here's why:

    The function in D is: f(x) = x^2 (1-x)^2

    Plugging in x=(1-x) into D gives:

    f(x) = (1-x)^2 (1-1+x)^2
    = (1-x)^2 x^2
    which is exactly the same function in D. Voila!


    Thank you so much for the help.

    I have another one if you dont mind, its to find the value for M.

    (1/5)^m*(1/4)^17=1/2(10^32)

    How do you find M with this abomination?


    Can you type it better? Is it this: 2(10^{32})
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    Jul 22, 2012 10:08 PM GMT
    ^^

    Yes that is what it is and I have no idea how to type like that...
  • Posted by a hidden member.
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    Jul 22, 2012 10:14 PM GMT
    Chainers said^^

    Yes that is what it is and I have no idea how to type like that...


    And do you have any idea what the answer is supposed to look like? Is it an exact number? Decimal etc? I'm trying to solve it exactly, but the answer isn't pretty, which makes me think that this shouldn't be right since your lovely standardised American tests aren't really like that.
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    Jul 22, 2012 10:18 PM GMT
    From what I remember the answer was a solid number.

    2 were in the 10's range, the other three were in the 30s.
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    Jul 22, 2012 10:20 PM GMT
    Chainers saidFrom what I remember the answer was a solid number.

    2 were in the 10's range, the other three were in the 30s.


    That could only mean the equation you provided is wrong. And this isn't my math skills speaking, it's an equation solver. According to the equation you gave, m≈)
    -60
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    Jul 22, 2012 10:26 PM GMT
    sirdreamalott said
    Chainers saidFrom what I remember the answer was a solid number.

    2 were in the 10's range, the other three were in the 30s.


    That could only mean the equation you provided is wrong. And this isn't my math skills speaking, it's an equation solver. According to the equation you gave, m≈)
    -60


    Can you tell me just how to go about solving the problem? I may be off by a couple of numbers but I have no idea how to approach exponential with 2 different bases.
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    Jul 22, 2012 10:35 PM GMT
    Chainers said
    sirdreamalott said
    Chainers saidFrom what I remember the answer was a solid number.

    2 were in the 10's range, the other three were in the 30s.


    That could only mean the equation you provided is wrong. And this isn't my math skills speaking, it's an equation solver. According to the equation you gave, m≈)
    -60


    Can you tell me just how to go about solving the problem? I may be off by a couple of numbers but I have no idea how to approach exponential with 2 different bases.


    That's the thing- if the problem is wrong it's kind of hard to have a good starting point. In any case, I would start off by taking the log of both sides and then using properties of logs to separate the terms.

  • Karl

    Posts: 5787

    Jul 22, 2012 11:52 PM GMT
    The equation (1/5)^m*(1/4)^17=1/2(10^32) has m ≈ - 60 , he's right .
    This is how I did it.
    There're something wrong with the equation you provided.
    2cbobj.jpg
  • Posted by a hidden member.
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    Jul 23, 2012 12:04 AM GMT
    Karl saidThe equation (1/5)^m*(1/4)^17=1/2(10^32) has m ≈ - 60 , he's right .
    This is how I did it.
    There're something wrong with the equation you provided.


    ^_^
  • Posted by a hidden member.
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    Jul 23, 2012 12:11 AM GMT
    So glad I'm done with this...
    This takes me back to my university days :-)
  • Posted by a hidden member.
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    Jul 23, 2012 12:36 AM GMT
    ermahgerd yer trorring ergern
  • Posted by a hidden member.
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    Jul 23, 2012 1:27 AM GMT
    I am happy to be a part of this asian persuasion. Thanks to all Asians who have helped us in our time of need.


    And please do not kill us all.
  • Posted by a hidden member.
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    Jul 23, 2012 1:46 AM GMT
    Mathematics that I used to knowwWWWwww

    Glee-608x367.jpg
  • Posted by a hidden member.
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    Jul 23, 2012 1:56 AM GMT
    METAMORPH saidMathematics that I used to knowwWWWwww

    Glee-608x367.jpg


    hahaha more like never.
  • Posted by a hidden member.
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    Jul 23, 2012 1:59 AM GMT
    Just give it up and go to law school.







    Just kidding. That all looks like gibberish to me, though. You boys is smart.
  • Karl

    Posts: 5787

    Jul 23, 2012 1:59 AM GMT
    METAMORPH saidMathematics that I used to knowwWWWwww

    {STIUTK pic}

    LOL
    me too , I wont have to see math for the rest of my life anymore icon_lol.gif