Question & Answer
QUESTION

The value of ${{(-2)}^{-2}}$is equal to.
 (a) $\dfrac{1}{2}$
 (b) $\dfrac{1}{4}$
 (c) $-\dfrac{1}{2}$
 (d) $-\dfrac{1}{4}$

ANSWER Verified Verified
Hint: We will use the laws of exponents. Exponents are better known as power or indices. The exponent of a variety says what percentage times to use the amount in multiplication.

Complete step-by-step solution -
We know that in an expression like ${{a}^{b}}$, a is called base and b is called power of the number.
Here, on comparing with ${{a}^{b}}$ we know that base is -2 and power is also -2.
Now according to law of exponents we know that: -
The exponents say how many times to use the number in multiplication.
A negative exponent means divides, because the opposite of multiplication is division.
A fraction exponent means taking root of the number.
Now since our power is negative, we will be using rule 2. For example, if we have ${{a}^{-b}}$ then this will evaluate to $\dfrac{1}{(a\times a\times a......btimes)}$. As ${{a}^{-b}}$ can be written as $1\times {{a}^{-b}}$ and therefore ${{a}^{-b}}$ is written as $\dfrac{1}{(a\times a\times a......btimes)}$.
So, according to rule 2 we will divide 1 by -2 two times.
 ${{(-2)}^{-2}}$= $\dfrac{1}{-2\times -2}$
We know that $-2\times -2=4$ because a negative number multiplied by a negative number of results in a positive number. So, the answer is
$\dfrac{1}{-2\times -2}$=$\dfrac{1}{4}$

Note: Now,if we talk about surds and indices there are several identities like
$\begin{align}
  & {{x}^{1}}=x \\
 & {{x}^{0}}=1 \\
 & {{x}^{m}}\times {{x}^{n}}={{x}^{m+n}} \\
 & {{({{x}^{m}})}^{n}}={{x}^{mn}} \\
 & {{x}^{-n}}=\dfrac{1}{{{x}^{n}}} \\
\end{align}$
Which can be remembered as a formula to solve the question quickly.