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This must be in the difficult pool question. Although seeming very easy to solve both statements, the question contains so many traps wait for all of those who are under pressure.

The only numbers that make this statement true are 0 and 1

Therefore 2) is insufficient.

So if we put the two together,
1) tells us x=2 or 0, but since 2) tells us x=1 or 0, then the combination of two results in only one answer which is 0.

I saw your posting ashkg, but needed to rephrase it this way to make sure I got the whole concept.

If I'm mistaken, somebody please let me know. _________________

What is the value of 0^0 (^ means exponent)? Can it be considered to be the limit of a^0 as a approaches 0 ?

0^0 is indeed indeterminate. It turns out that you could make it have any value between 0 and 1, inclusive. You could have 0 if it's the limit as a->0 of 0^a, you could have 1 if it's the limit as a->0 of a^0, and for x in between 0 and 1, (and this is the neat part from Dr. Shimimoto) look at the expression (x^n)^(1/n). This just equals x for all positive values of n. As n->Infinity, this fraction goes to 0^0, but if it's just x the whole time, the limit of the expression as it goes to 0^0 is x. So we could make it anything in between 0 and 1, so it's got to be indeterminate.

Some technical crap...

In sum, I think we can't take x=0 as one of the solution.
So, the answer should be (A)..

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...