D?
Taking 0 not as an option as ETS considers 0^0 undefined. So, +1,-1,+2,-2 ?

I created this question to remind students that, when it comes to equations with variables in the exponents, there are three important provisos we must consider before we can conclude that two exponents are equal.
That is, if \(x^a = x^b\), then we can conclude that \(a = b\) AS LONG AS \(x \neq 0\), \(x \neq 1\), and \(x \neq -1\).
For example, if we know that \(0^x = 0^3\), we can't then conclude that \(x = 3\), since there are infinitely many values of \(x\) that satisfy the equation.

Let's begin by assuming \(b\) does not equal any of the forbidden numbers (i.e., \(x \neq 0\), \(x \neq 1\), and \(x \neq -1\))
In that case, we can conclude that \(2b^2 = 8\)
Divide both sides of the equation by \(2\) to get: \(b^2 = 4\)
So, \(b = 2\) or \(b = -2\)

Now we need to check whether each of the forbidden numbers is a possible solution to the given equation.

We already said that b is a non-zero integer, so we need not bother testing \(b = 0\).

What about \(b = 1\)?
Plug it in to get: \(1^{(2(1^2))} = 1^{8}\)
Simplify: \(1^{2} = 1^{8}\)
Works!!
So, \(b = 1\) is a solution

What about \(b = -1\)?
Plug it in to get: \((-1)^{(2(-1)^2)} = (-1)^{8}\)
Simplify: \((-1)^{2} = 1^{8}\)
Works!!
So, \(b = -1\) is a solution

So we have a total of four solutions: \(b = 2\), \(b = -2\), \(b = 1\) and \(b = -1\)

Answer: D

If you want to practice this question type, here’s a similar question:
https://gre.myprepclub.com/forum/what-is-s ... 20887.html

GreenlightTestPrep
You are right

Posted from my mobile device

Hi Brent GreenlightTestPrep, a bit confused that after we found out b = 2 or -2, not sure why are we start testing with 1 and -1 (forbidden numbers not allowed) instead of 2 and -2? Thanks Brent

You need to find out the solutions for which in the end we do have b^8

2 and -2 are one pair

Then you use 1 and -1

Seee more theory here https://gre.myprepclub.com/forum/gre-ma ... 29264.html

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