Courses for Kids
Free study material
Offline Centres
Store Icon

How do you simplify ${{\left( \dfrac{1}{16} \right)}^{\dfrac{-3}{4}}}$?

Last updated date: 16th Jul 2024
Total views: 381.6k
Views today: 5.81k
381.6k+ views
Hint: Here an expression is given which has a fraction exponent.
For solving fraction exponent, we have to use some general rules of fractional exponent.
As for ${{n}^{th}}$ root and $m$ power for number $a$, the general form can be written as,
\[{{a}^{\dfrac{m}{n}}}={{\left( \sqrt[n]{a} \right)}^{m}}\]
\[n=\] root
\[m=\] power
By using the above fraction rule we can solve the given expression.

Complete step by step solution:Given that, there is an expression having fractional exponent as,
${{\left( \dfrac{1}{16} \right)}^{\dfrac{-3}{4}}}$, and we have to solve it,
We know the same general rules for fraction exponent which is as follows,
And \[{{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}\]
Now using above general rules we have to solve the given expression
${{\left( \dfrac{1}{16} \right)}^{\dfrac{-3}{4}}}$ (given)
Now use the first rule which we have already written,
i.e. \[{{x}^{-1}}=\dfrac{1}{x}\]
\[\therefore {{\left( \dfrac{1}{16} \right)}^{\dfrac{-3}{4}}}={{\left( 16 \right)}^{\dfrac{3}{4}}}\]
But, we know that, \[{{\left( 2 \right)}^{4}}=16\]
So, now use the second rule which we have written
i.e. \[{{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}\]
\[\therefore {{\left( {{2}^{4}} \right)}^{\dfrac{3}{4}}}={{2}^{{4}\times \dfrac{3}{{{4}}}}}\]
And the expression becomes
\[{{\left( \dfrac{1}{16} \right)}^{\dfrac{-3}{4}}}={{2}^{3}}\,\,=\,8\]

Additional Information:
Expressing the power and roots together is known as fractional exponent.
The general form of writing the fractional exponent for \[{{n}^{th}}\] root is as follows:
\[\therefore {{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]
It means that, when \[{{n}^{th}}\] root of \[a\] is multiplied by \[n\] times, it will give the result as \[a\]
\[{{a}^{\dfrac{1}{n}}}\times {{a}^{\dfrac{1}{n}}}\times {{a}^{\dfrac{1}{n}}}\times .......\times {{a}^{\dfrac{1}{n}}}=a\]
For example \[{{625}^{\dfrac{1}{4}}}\]
\[\therefore {{625}^{\dfrac{1}{4}}}=\sqrt[4]{625}\]
As we know that, \[{{5}^{4}}=625\]
Therefore, the answer will be
In the above equation \[625\]is radicand as it is under the radical sign.
The order of the above equation is \[4\] as it indicates the root. Now, for \[{{n}^{th}}\] root and \[m\]power of the general form ca =n be \[{{a}^{3}}\]
\[{{a}^{\dfrac{m}{n}}}={{\left( \sqrt[n]{a} \right)}^{m}}\]
It means that we have to take the root of \[a\] as \[n\] and power as \[m\] as, \[{{\left( {{a}^{\dfrac{1}{n}}} \right)}^{m}}\]

In this numerical, the given expression has a fractional exponent. So, we have to solve it by using the rules of fraction exponent. The given expression can solve in other way also which is as follows
${{\left( \dfrac{1}{16} \right)}^{\dfrac{-3}{4}}}$ given
But as we know that,
\[\therefore {{\left( \dfrac{1}{16} \right)}^{\dfrac{-3}{4}}}={{(16)}^{\dfrac{3}{4}}}\]
Now according to general form of \[{{n}^{th}}\] root and \[m\] power for number \[a\] is,
\[{{a}^{\dfrac{m}{n}}}={{\left( \sqrt[n]{a} \right)}^{m}}\]
So, from above general form compare to \[{{(16)}^{\dfrac{3}{4}}}\] we have \[4\] as a root and \[3\] as a power
i.e. \[n=4;\,m=3\]
Therefore, the equation becomes,
\[{{\left( 16 \right)}^{\dfrac{3}{4}}}={{\left( \sqrt[4]{16} \right)}^{3}}\]
As we know that, \[{{2}^{4}}=16\]
Therefore, \[{{\left( 16 \right)}^{\dfrac{3}{4}}}={{\left( \sqrt[4]{24} \right)}^{3}}\]
\[={{\left( 2 \right)}^{3}}=8\]
Therefore, the final solution for given expression \[{{\left( \dfrac{1}{16} \right)}^{\dfrac{-3}{4}}}\] is \[8\]