Exponents: Is it true that x > 0 ?

from question we can we can say that x^2 = 1;

so answer should be d.

Here's what I have
C

1) x = x^2-2
or.... 0 = x^2-x-2
0 = (x-2) (x-1)
x = 2 or x = -1 ........Not Suff.

2) 2x<X^5
0<x^5-2X
0<(x) (X^4 - 2)
0<x
2<X^4..........Not Suff.

C) Since 2<X^4..........
Only 2^4>0....................Suff.
(-1)^4 < 0

Please correct me if I'm wrong on this...

For Prob. 1

1) x = x^2-2
or.... 0 = x^2-x-2
0 = (x-2) (x-1)
x = 2 or x = -1 ........Not Suff.


I meant 0 = (x-2) (x+1)....Not 0 = (x-2) (x-1)

IMO A.
What os OA

The answer should be D.

x^n=x^n * x^2

x^2 = 1
x = +1 or -1

Statement I

x=x^2 - 2

if x =-1
x = -1

if x=1
x=-1

Sufficient.

Statement II
2x < x^5

if x =-1
-2 < -1

if x=1
2<1 which is not possible.
Therefore x=-1

Sufficient.

Hence D is the answer

OA is D.

This is testing your knowledge of exponents ... and how well you know your exponent rules.
Remember the rule for exponents:
An exponent expression with a base of 0 will always yield 0, regardless of the exponent
And exponent expression with a base of 1 will always yield 1, regardless of the exponent.
So if you are told that x^6 = x^7 = x^15, so x must be 0 or 1.
If you are told x^6 = x^8 = x^10, x could be 0, 1, or -1 (note even exponents here)

Lets take a look at the Q-stem:
x^n = x^n+2
What is this telling us.
This is telling ust that x could be 0, 1, or -1.
0^n will always = 0^n+2
1^n will always = 1^n+2
(-1)^n will always = (-1)^n+2 (Note: if n is even, then n+2 is also even. If n is odd, then n+2 is also odd.)

So from the Q-stem we know that the possible values of x are 0, 1, -1.

(1) x = x^2 - 2
=> 0 = x^2 + x - 2
=> 0 = (x+1)(x-2)
So x = -1 or x = 2.

We know from Q-stem that x could be 0, 1, or -1.
x = 2 is not in this list. Only -1 is. so (1) is SUFF.

(2) 2x < x^5
We use the possible values for x from q-stem (0, 1, -1) and plug them into this inequality, to see which will make this inequality true.

Lets try -1:
2(-1) < (-1)^5
-2 < -1
this is true so x could be -1

Lets try 0:
2(0) < (0)^5
0 < 0
this is NOT true, so x cannot be 0

Lets try 1:
2(1) < 1^5
2 < 1
this is NOT true, so x cannot be 1.

Only poss value of x is -1. So (2) is SUFF.

Hence answer D.

I think if you remember the exponent rules ... then it sets you off on the right foundation in answering this question.

I was really interested in finding out alternative ways in solving this question. Did anyone use a different approach in solving this ?

Thanks.

Are you sure for this one ? "0^n will always = 0^n+2"
if n=0 0^0=1 and 0^2=0

pepeprepa wrote:Are you sure for this one ? "0^n will always = 0^n+2"
if n=0 0^0=1 and 0^2=0

Good question.
My understanding is that ... Where x is not equal to 0. x^0 will always be 1. E.g. 6^0 = 1
However ... putting aside exponent rules for one minute ... lets consider what 0^0. The base here is 0 ... this mean there is nothing there in the first place ... so how can we get 1 from 0 !

Maybe someone else has a better explanation on this point.

Thanks.

You are right 0^0 does not exist.. I have vague remembers that we used it during our classes, perhaps kind of tips for some situations, but we cannot state that 0^0 that's WRONG.

pepeprepa wrote:You are right 0^0 does not exist.. I have vague remembers that we used it during our classes, perhaps kind of tips for some situations, but we cannot state that 0^0 that's WRONG.

yes ... I think sometimes you have to take a step back from all the math rules and apply some common sense to these types of items.

It is a convention, I don't think you can see it with good sens. You can consider it in some context/formula (so it can be logical) but you cannot state it in some other contexts... You need to know when you can use it, for gmat it is useless.