If wxy ≠0, does x = y ?
(1) w^x = w^y
(2) wxy ≠xy
I need some clarification on stm1....
If WXY does not = 0
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Hi,
Stmt 1:
Taking logarithm on both sides;
log (w^x) = log (w^y)
Using power rule,
x times log (w) = y times log (w)
* Edited once *
Dividing throughout by log(w) can be done only if it is not equal to 0. log(w) will become 0 if w equals 1.
Hence, we are not able to divide on both sides by log(w) in the absence of additional information about w.
Thanks.
Stmt 1:
Taking logarithm on both sides;
log (w^x) = log (w^y)
Using power rule,
x times log (w) = y times log (w)
* Edited once *
Dividing throughout by log(w) can be done only if it is not equal to 0. log(w) will become 0 if w equals 1.
Hence, we are not able to divide on both sides by log(w) in the absence of additional information about w.
Thanks.
Last edited by 4GMAT_Mumbai on Thu Dec 02, 2010 10:29 pm, edited 1 time in total.
Naveenan Ramachandran
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4gmat i was thinking same like you except i used the rule of exponent for like bases. However according to the solution we are both wrong..
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Hi ...
My bad ... The logic for statement 1 falls apart when log (w) = 0; that is, if w = 1.
Statement 2 gives the info that w is not equal to 1.
Is the answer C by any chance ...
My bad ... The logic for statement 1 falls apart when log (w) = 0; that is, if w = 1.
Statement 2 gives the info that w is not equal to 1.
Is the answer C by any chance ...
Naveenan Ramachandran
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Given: wxy ≠0HPengineer wrote:If wxy ≠0, does x = y ?
(1) w^x = w^y
(2) wxy ≠xy
Implies none of w, x and y is equal to zero.
Statement 1: w^x = w^y
As none of w, x and y is zero this statement implies,
- (1) w = 1, x and y can be any number.
(2) w = -1, x and y are both even or both odd integers.
Statement 2: wxy ≠xy
As x and y not zero this statement implies w ≠1 and nothing else.
Not sufficient.
1 & 2 Together: w must be equal to -1 and in that case x and y have to be both even or both odd integers (not necessarily same integer).
Not sufficient.
The correct answer is E.
Last edited by Rahul@gurome on Sat Dec 04, 2010 7:05 am, edited 1 time in total.
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@ Rahul,1 & 2 Together: w must be equal to -1 and in that case x and y can be any even integers (not necessarily same integer).
Not sufficient.
Why should the x and y be any even integer.
Infact it can be any integer.
By the conditions given,
1> w^x = w^y
Here w can be one, and x and y can be any positive integer.
2> wxy is not equal to xy..=> w is not equal to 1.
In essense we find w not equal to 0 and 1.
So w can be -1.
1> w^x = W^y
(-1)^1 = (-1)^3
We get x as 1 and y as 3. But infact x and y also can be the same to yield the same answer.
X and y can be anything and not just any even integer.
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x and y can't be any integer, they have to be both even or both odd for w = -1. If x is odd and y is even then w^x = -1 and w^y = 1, which are not equal. Same for x even and y odd.junegmat221 wrote:@ Rahul,
Why should the x and y be any even integer.
In fact it can be any integer.
...
X and y can be anything and not just any even integer.
But there was a mistake. I forgot to mention the odd integer. Edited the reply.
Thanks.
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