Question

How do you simplify ${{\left( \dfrac{49}{81} \right)}^{-\dfrac{1}{2}}}$?

Answer 1

Hint: It can be simplified by making certain transformations and substitutions. Negative powers indicate how many times we divide the number. Here the negative power is in fractional form, so we have to approach very carefully to simplify it. We know that if the power is in fractional form and the fraction part is $\dfrac{1}{2}$, then it indicates that it is the square root of the base number. But from the given question we can see that the exponential part is in the form of a negative fraction that is $-\dfrac{1}{2}$. It indicates that we divide the number by $-\dfrac{1}{2}$ times. That is one by square root of the base part.

Complete step by step answer:
From the question it had been given that to simplify ${{\left( \dfrac{49}{81} \right)}^{-\dfrac{1}{2}}}$
Hence, the simplification is
As we have been already discussed above the simplification follows like below,
${{\left( \dfrac{49}{81} \right)}^{-\dfrac{1}{2}}}=\dfrac{1}{\sqrt{\dfrac{49}{81}}}$
We know that the square root of $49$ is $7$
And we also know that square root of $81$ is $9$
Now, using the above known square roots, on further simplification we get the below simplified form,
${{\left( \dfrac{49}{81} \right)}^{-\dfrac{1}{2}}}=\dfrac{1}{\dfrac{7}{9}}$
$\Rightarrow {{\left( \dfrac{49}{81} \right)}^{-\dfrac{1}{2}}}=\dfrac{9}{7}$
Hence the simplified form of given ${{\left( \dfrac{49}{81} \right)}^{-\dfrac{1}{2}}}=\dfrac{9}{7}$
Hence the given question is simplified.

Note: We should be very careful while handling with the negative exponents and negative fractional exponents. We should be well aware of the simplification of the negative fractional part problems. We also should be well aware of the square roots. This can also be done as
 $\begin{align}
  & {{\left( \dfrac{49}{81} \right)}^{-\dfrac{1}{2}}}={{\left( \dfrac{{{7}^{2}}}{{{9}^{2}}} \right)}^{-\dfrac{1}{2}}} \\
 & \Rightarrow {{\left( \dfrac{7}{9} \right)}^{\left( 2 \right)\left( -\dfrac{1}{2} \right)}}={{\left( \dfrac{7}{9} \right)}^{-1}} \\
 & \Rightarrow \dfrac{9}{7} \\
\end{align}$