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Official Solution



Given:

    • \(\sqrt{x}=y\)

Working Out:

    • \(\frac{1}{(x^{−2})^{−2}}\)

    • \(\frac{1}{x^4}\)

It is given that

    • \(\sqrt{x}=y\)

Taking 8th power on both sides we get:

    • \(x^4 = y^8\)

Therefore,

    • \(\frac{1}{x^4} =\frac{1}{y^8} =y^{-8}\)

Correct Answer: Option A


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EgmatQuantExpert
Q.
If \(\sqrt{x}=y,\) which of the following could be equal to \(\frac{1}{(x^{−2})^{−2}}\)

Answer Choices

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

Thanks,
Saquib
Pick nice numbers, and y is positive. When the principal square root symbol is written in the problem, we need to consider the positive square root only.

\(\sqrt{x}=y\), and \(\sqrt{4}= 2\), so x = 4 and y = 2

\(\frac{1}{(x^{−2})^{−2}}\) = \(\frac{1}{x^4}\) ---> substitute 4 for x

= \(\frac{1}{4^4}\)

= \(\frac{1}{(2^{2})^{4}}\)

= \(\frac{1}{2^8}\)

= \(2^{-8}\)

Substitute y for 2 ---> \(y^{-8}\)

Answer A

Shorten the math: \(\frac{1}{4^4}\) = \(\frac{1}{256}\), and \(\frac{1}{256}\) = \(\frac{1}{2^8}\), which is \(2^{-8}\). Substitute y for 2 ---> \(y^{-8}\)
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Q.

If \(\sqrt{x}=y,\) which of the following could be equal to \(\frac{1}{(x^{−2})^{−2}}\)

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

\(\sqrt{x}=y,\)

Or, \(x = y^2\)

Quote:
which of the following could be equal to \(\frac{1}{(x^{−2})^{−2}}\)

\(\frac{1}{(x^{−2})^{−2}}\)

= \(\frac{1}{(x^{−4})}\)

= \(x^4\)

= \(y^{2*4}\)

= \(y^{8}\)


Hence, answer will be (A) \(y^{8}\)
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I consider it very easy EGMAT is rarely kind for students
Answer:A
Time consumed:0:50(I cross check since was in doubt of EGMAT kindness)
√x=y
x^1/2=y
we have to make it x^4
So x^1/2*8=y^8
Multiply both sides with 8
Sp,
X^4=y^8
Since
1/(x^-2)^-2
1/x^4
put equivalent of x^4
it is
1/y^8
which is
y^-8
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