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Great catch - missed that if x=0, a and b would have to be positive.

But even if we take the two statements together, couldn't

1. x=0, (a,b) = (2,3)
This would satisfy x^a = x^b as well as x|a|=x|b|
or
2. x=1, (a,b) = (-4,4)
This would satisfy x^a = x^b as well as x|a|=x|b|

So shouldn't it be E?
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You are right :) I missed it.

in stmt2: we cannot cancel out x until we know that x is not equal to zero.

E should be the answer.
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Think about this way if x = 1 or 0

then if a = 1 and b= -1

both statements are satisfied but a not = b
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Think about this way if x = 1 or 0

then if a = 1 and b= -1

both statements are satisfied but a not = b

Thats correct. OA: Key here is to think about x also and run it through the zero, integer, negative test. And as sher676 said, for x=1 and x=0 there maybe many combination of a & b!

Economist, you got that right.. I actually was looking for a while how to put a "crossed=" :)

We can use neq with the math button \(ab \neq 0\)
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Thats a gud 1 gmattokyo, (zero, integer and negative test) - if this test is used many DS ques can be answered correctly. Thanks for tht tip! Going by this the [E] has to be the answer..
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good question.. not difficult but a little tricky!
thanks for your constant supply of questions gmattokyo.. providing some good practice!
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good question.. not difficult but a little tricky!
thanks for your constant supply of questions gmattokyo.. providing some good practice!

Thanks sriharimurthy. glad ur finding it of help :) lets crack GMAT (and try remain sane at the same time)!
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sriharimurthy
good question.. not difficult but a little tricky!
thanks for your constant supply of questions gmattokyo.. providing some good practice!

Thanks sriharimurthy. glad ur finding it of help :) lets crack GMAT (and try remain sane at the same time)!


Haha. Cheers to that gmattokyo!
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gmattokyo
if ab <> 0, is a=b?

1) \(x^a = x^b\)
2) \(x|a|=x|b|\)

I'm a bit suprised to find that the answer is (E)
Statement 1 is clearly insuff
Statement 2 though x|a|=x|b|, so |a|/|b| = 1. Hence since the answer is positive, both need to be negative or both need to be positive.
In any case a = b.

I wonder what am I doing wrong here.
Experts please advice

Thank you
Cheers
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gmattokyo
if ab <> 0, is a=b?

1) \(x^a = x^b\)
2) \(x|a|=x|b|\)

I'm a bit suprised to find that the answer is (E)
Statement 1 is clearly insuff
Statement 2 though x|a|=x|b|, so |a|/|b| = 1. Hence since the answer is positive, both need to be negative or both need to be positive.
In any case a = b.


I wonder what am I doing wrong here.
Experts please advice

Thank you
Cheers

Hi,
|a| and |b| has to be positive; they can not be negative since they are absolute values. Moreover when we have |a| = |b|; a and b can be positive/negative

for eg. a=2 b=-2 ; Therefore |a| = |b| = 2.

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gmattokyo
if ab <> 0, is a=b?

1) \(x^a = x^b\)
2) \(x|a|=x|b|\)

I'm a bit suprised to find that the answer is (E)
Statement 1 is clearly insuff
Statement 2 though x|a|=x|b|, so |a|/|b| = 1. Hence since the answer is positive, both need to be negative or both need to be positive.
In any case a = b.


I wonder what am I doing wrong here.
Experts please advice

Thank you
Cheers

There are two problems:
1. We cannot reduce x|a|=x|b| by x to get |a|=|b|. Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

So when you divide by x you assume, with no ground for it, that x does not equal to zero thus exclude a possible solution. Notice that x=0 and |a|=|b| both satisfy x|a|=x|b|.

2. Even if we had |a|=|b|, it would still be insufficient. |a|=|b| means that a=b or a=-b (consider a=b=1 and a=-b=1).

Hope it helps.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


If ab ≠ 0, does a=b?

(1) x^a = x^b
(2) x|a| = x|b|


In the original condition, there are 2 variables(a,b). But, overall, there are 3 variables(a,b,x), which should match with the number of equations. So you need 3 equations. For 1) 1 equation, for 2) 1 equation, and you lack 1 equation, which is likely to make E the answer. When 1) & 2), if x=1, a=b=2 -> yes, a=2, b=-2 -> no, which is not sufficient. Therefore, the answer is E.


 For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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gmattokyo
If ab ≠ 0, does a=b?

(1) x^a = x^b
(2) x|a| = x|b|

Since ab ≠ 0, that means neither a nor b is 0. We need to determine whether a = b.

Statement One Alone:

x^a = x^b

If x = 2, then a = b. However, if x = 1, then a and b could be any numbers. For example, a could be 1 and b could be 2.

Statement one alone is not sufficient.

Statement Two Alone:

x|a| = x|b|

We see that a could equal b; for example, x = 1 and a = b = 1. However, a could also not equal b, for example, x = 1, a = 1 and b = -1.

Statement two is not sufficient.

Statements One and Two Together:

With the two statements, we still can’t determine whether a = b.

For example, x = 1 and a = b = 1 or x = 1, a = 1 and b = -1 and thus a does not equal b.

Answer: E
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Hi, can someone pls explain why statement 1 is insufficient. I can't seem to understand the logic.
Thanks!
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Hi, can someone pls explain why statement 1 is insufficient. I can't seem to understand the logic.
Thanks!
Statement 1:
x^a=x^b

If X is not -1,0,1, then a must be equal to b
If X is -1,0,1 then a or b can take any value. For example, 1^100=1^2.


As we cannot determine the value that X takes, or a range of values, it is insufficient.

Note that even if, for example, we know that X=1, it is still insufficient since a and b can assime any value.



Posted from my mobile device
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Oh ok I got it now! Thanks a ton!
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Bunuel please help me to see the mistake in statement 1

x^a=x^b
x^a/x^b=1
x^a-b=1
x^a-b=x^0
a-b=0
a=b
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