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if x and y are intigers is x^y * y^-x = 1?? 1) x^x > y 2) [#permalink ]
16 Sep 2006, 03:59
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This topic is locked. If you want to discuss this question please re-post it in the respective forum. if x and y are intigers is x^y * y^-x = 1??
1) x^x > y
2) x>y^y
Director
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I used following
A) X^X=Y then stem is =1 but since it is inequality, then it is not = to 1
IMO it should be D
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For me, it's (B)
Statment 1:
If x = y = 2, then :
> 2^2 = 4 > 2
> 2^2 / (2^2) = 1
If x = 2 and y = 1, then :
> 2^2 = 4 > 1
> 2^1 / (1^2) = 2
INSUFF
Statment 2:
x > y^y thus x > y (x & y are integers) that means x != y
Hence, x^y * y^-x != 1
SUFF
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Re: Excellent iniquality DS [#permalink ]
16 Sep 2006, 06:23
yezz wrote:
if x and y are intigers is x^y * y^-x = 1?? 1) x^x > y 2) x>y^y
Can some one see if my working is correct?
given x,y are ints
is x^y * y^-x = 1
or x^y / y^x = 1 ===> x^y = y^x
taking logs on both sides we have to prove (y/x)(logx/logy)=1
from (1) we have
xlogx > log y
logx/logy > 1/x
(so we do not know anything about it)
from (2) we have
x>y^y ==> logx > ylogy
Means
logx/logy > y
Then I am lost...
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Hallo trivikram
think that ; or x^y / y^x = 1 ===> x^y = y^x
is NOT correct .It should be (x^y-y^x)/y^x=0 where y^x can not be 0 since division by 0 is not defined
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BG wrote:
Hallo trivikram think that ; or x^y / y^x = 1 ===> x^y = y^x is NOT correct .It should be (x^y-y^x)/y^x=0 where y^x can not be 0 since division by 0 is not defined
But BG from there how do we proceed?
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Re: Excellent iniquality DS [#permalink ]
16 Sep 2006, 07:37
yezz wrote:
if x and y are intigers is x^y * y^-x = 1?? 1) x^x > y 2) x>y^y
The condition x^y * y^-x = 1 will be valid only if x and y are 2 and 4
from statement 2 we can say that, that is not the case. Hence B is the anwer.
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.........................
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mailtheguru wrote:
.........................
What's going on ?
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for me B is sufficient...
X and Y are integers... we dont know if they are postive or negative...
x^y * y^-x=1?
(1) x^x>y
well say if x and y=2...then 2^2>2...
2^2*1/(2^2)=1; x=2, y=3 then the stem isnt equal to 1
(2)
x>y^y; so we know that |x|>|y| and they are not equal..therefore the stem will never be 1....
sufficient
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Fig wrote:
mailtheguru wrote:
.........................
What's going on ?
Got screwed
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mailtheguru wrote:
Fig wrote:
mailtheguru wrote:
.........................
What's going on ?
Got screwed
It's one part of the joy of Maths (or DS?
).... One time u another time someone else (this rule works with me as well
)
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Fig wrote:
mailtheguru wrote:
Fig wrote:
mailtheguru wrote:
.........................
What's going on ?
Got screwed
It's one part of the joy of Maths (or DS?
).... One time u another time someone else (this rule works with me as well
)
yeah, sure.
Where are these questiosn from?
Director
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A very shaky B for me as well...
x^y = y^x
=> ylogx = x logy
=> y/x = logy/logx -- (1)
A# x^x > y
=> xlogx > logy
=> From (1), y/x^2 = logy/xlogx < 1
=> y < x^2
Now, if x=2, we have y = 1, 2, 3... if y = 2 then x^y = y^x, but not in the case of y =1 or 3. Hence not sufficient.
B# x > y^y
=> logx > ylogy
=> from (1), y^2/x = ylogy/logx < 1
=> y^2 < x
Now, if x = 27, we have y = 1, 2, 3, 4 or 5; but the expression in the question does not evaluate to 1 in any of the cases. Hence sufficient.
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if x and y are intigers is x^y * y^-x = 1??
1) x^x > y
2) x>y^y
rephrase stem x^y *1/y^x = 1 ie: receprocals
This is only possible if x^y = y^x and in turn this is only possible if x=y or x,y belongs to the set {2,4}
from one
x^x > y thus sure x,y are not equall one
x,y could be anything and could be x=4 and y=2 .....insuff
from two
x>y^ythus sure x,y are not equall one and sure /x/> /y/ thus if they are 2,4 x must be the 4 and thus y^y = 4....suff
because now we are sure that the only two conditions that make the assumption true are not valid ( being one or {2,4})
suff
answer is not a shaky b it is B
Last edited by
yezz on 16 Sep 2006, 12:35, edited 2 times in total.
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yezz wrote:
if x and y are intigers is x^y * y^-x = 1??
1) x^x > y
2) x>y^y
rephrase stem x^y *1/y^x = 1 ie: receprocals
This is only possible if x^y = y^x and in turn this is only possible if x=y=1 or x,y belongs to the set {2,4} from one
x^x > y thus sure x,y are not equall one
x,y could be anything and could be x=4 and y=2 .....insuff
from two
x>y^ythus sure x,y are not equall one and sure /x/> /y/ thus if they are 2,4 x must be the 4 and thus y^y = 4....suff
because now we are sure that the only two conditions that make the assumption true are not valid ( being one or {2,4})
suff
answer is not a shaky b it is B
Not agree with the bold
x = y makes it
and no restricted values on it apart 0^0 (recently identified as not fitted for GMAT
)
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come on Fig you ve got to get over this exponent DS ...
and i agree with you on your comment but still it would be insuff
I will edit it
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It's the end of my day here
I'm allowed to turn the end of my post to something funny for those who know it
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Man GMAT is funny
funny wn we do silly mistakes " kinder garten ones" apart from the test
and funny when you laugh yourself to death infront of a hard DS that you ve no clue about
we need those jokes all the time
by the way i like your audi
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Ahhh
If u start to speak about Audi... I'm not ready to make an end to this day finally
Frankly, I recommand this brand over the 2 other major german brands
But I try to convince no one
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