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If WXY does not = 0

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Source: — Data Sufficiency |
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by 4GMAT_Mumbai » Thu Dec 02, 2010 10:09 pm
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.
Last edited by 4GMAT_Mumbai on Thu Dec 02, 2010 10:29 pm, edited 1 time in total.
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by HPengineer » Thu Dec 02, 2010 10:13 pm
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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by 4GMAT_Mumbai » Thu Dec 02, 2010 10:27 pm
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 ...
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by HPengineer » Thu Dec 02, 2010 11:48 pm
official answer is E...
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by Rahul@gurome » Fri Dec 03, 2010 12:14 am
HPengineer wrote:If wxy ≠ 0, does x = y ?

(1) w^x = w^y

(2) wxy ≠ xy
Given: wxy ≠ 0
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.
Not sufficient.

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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by 4GMAT_Mumbai » Fri Dec 03, 2010 10:13 pm
Respect ! <Bow> !!
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by junegmat221 » Sat Dec 04, 2010 6:32 am
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.
@ Rahul,
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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by Rahul@gurome » Sat Dec 04, 2010 7:11 am
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.
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.

But there was a mistake. I forgot to mention the odd integer. Edited the reply.

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