Question

How do you simplify \[{{\left( 4xy \right)}^{-2}}\] ?

Answer 1

Hint: For problems like these we need to recapitulate the theory of power and indices and we also need to have a fair bit of idea regarding factorization. We need to know the general representation of a number in its power form. After that, we need to represent our problem in the simplest way possible. We know that in cases of multiplication of numbers or variables the associative law holds good, and because of this, solving the problem becomes easy. We then just need to find the value of the power for each and every individual term inside the bracket. We know that any power form of a number can be represented as \[{{a}^{b}}\] . Here, ‘a’ is called the “base” and ‘b’ is called the “index” or “power” or “exponent”.

Complete step by step solution:
Now, we start off with our solution by writing that, we first off apply the associativity property to our given problem sum and it forms,
\[\begin{align}
  & {{\left( 4xy \right)}^{-2}} \\
 & \Rightarrow {{4}^{-2}}{{x}^{-2}}{{y}^{-2}} \\
\end{align}\]
Now, in this problem we convert the negative power of the bases to positive power by taking the reciprocals. We write it as,
\[\Rightarrow \dfrac{1}{{{4}^{2}}{{x}^{2}}{{y}^{2}}}\]
Evaluating the values we write,
\[\Rightarrow \dfrac{1}{16}\dfrac{1}{{{x}^{2}}{{y}^{2}}}\]
We can observe from this above equation that the equation cannot be further simplified, hence this is our answer to the problem.

Note:
For problems like these, where we need to find the square or cube of any number or any variable, we must be very careful with the conversion of negative powers to positive powers and we must always note that the \[{{a}^{-n}}th\] power is equivalent to \[\dfrac{1}{{{a}^{n}}}th\] power of a number or a variable. We must also be careful with the properties, and should not overlap one property with the other. We should also be careful while applying the associative property or any other laws of indices or factorization.