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If x is an integer, how many possible values of x satisfy the equation [#permalink]
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The only number with an exponent or elevated whose result is 1, regardless, is when a number has zero as the exponent or the base itself is one. 1 raised to the power of 100 is still 1

Not only that: notice how the exponent is even because if you have 2x+2

x= 0 >>> even
x=1 >> even
......

And because the exponent is even always, we do know that when \(x^2\) our base or \(x\) could be also negative

\(1^2=1\)
\(-1^2=1\)

still positive for any number

So we must have two scenarios at the same time

1) the power is zero , so the number raised to zero is 1

OR

2) a number, one positive and one negative, raised to the power, regardless the entity of the power, you still get 1



The three values are: zero, one, and minus one \(= 0,1,-1\).

So three values and the answer is 3


Tough question , really.

You need to master as cold the exponent and number properties rules

See the new quant handout. I have just completed it

https://gre.myprepclub.com/forum/gre-math- ... tml#p95352
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