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JcAlmighty

The answer is E.4

The equation is: \(n ^{-2}= \frac{1}{16}\)

Since \(x^{-a}= \frac{1}{ x^{a} }\), then \(n^{-2} = \frac{1}{n^{2} }\)

Therefore, the equation is:
\(\frac{1}{ n^{2} } = \frac{1}{16}\)
⇒ \(n^{2} = 16\)
    \(n= \sqrt{16}=4\)

Answer:

Option B and E are correct

n could be 4 and -4

Step-by-step explanation:

Given the following:

\(n^{-2} = \frac{1}{16}\)

we know that:

\(a^{-n} = \frac{1}{a^n}\)

Then;

\(\frac{1}{n^2}=\frac{1}{16}\)

By cross multiply we have;

\(16=n^2\)

or

\(n^2=16\)

⇒\(n = \pm\sqrt{16}\)

Simplify:

\(n =\pm 4\)

⇒n = -4 and n = 4

Therefore, n could be 4 and -4

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