Problem Solving

If $$x=\sqrt{y},$$ which of the following could be equal to $$\frac{1}{\left(x^{-2}\right)^{-2}}\ ?$$

$$A.\ y\ ^{-8}$$
$$B.\ y\ ^{-4}$$
$$C.\ y\ ^{-2}$$
$$D.\ \ y\ $$
$$E.\ \ y\ ^2$$

The OA is A .

How can I show that A is the correct answer? Can any Expert help me?

Hello Vjesus12.

Let's take a look at your question. $$If\ \ \ \sqrt{x}=y,\ then\ x=y^2\ and\ then\ x^4=\left(y^2\right)^4=y^8.$$ By the other hand, $$\left(x^{-2}\right)^{-2}=x^4.$$ So, $$\frac{1}{\left(x^{-2}\right)^{-2}}=\frac{1}{x^4}=\frac{1}{y^8}=y^{-8}.$$

So, the correct answer should be A .

I hope this explanation may help you.

Feel free to ask me again if you have a doubt.

EconomistGMATTutor wrote:Hello Vjesus12.

Let's take a look at your question. $$If\ \ \ \sqrt{x}=y,\ then\ x=y^2\ and\ then\ x^4=\left(y^2\right)^4=y^8.$$ By the other hand, $$\left(x^{-2}\right)^{-2}=x^4.$$ So, $$\frac{1}{\left(x^{-2}\right)^{-2}}=\frac{1}{x^4}=\frac{1}{y^8}=y^{-8}.$$

So, the correct answer should be A .

I hope this explanation may help you.

Feel free to ask me again if you have a doubt.

Some correction in order. The question states that X = Y^1/2.

So 1/X^4 = X^-4 = Y^-2 C