Answer

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**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,

\[{{a}^{\dfrac{m}{n}}}={{\left( \sqrt[n]{a} \right)}^{m}}\]

Where,

\[n=\] 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,

\[{{x}^{-1}}=\dfrac{1}{x}\]

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. \[{{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}\]

\[\therefore {{\left( {{2}^{4}} \right)}^{\dfrac{3}{4}}}={{2}^{{4}\times \dfrac{3}{{{4}}}}}\]

\[={{2}^{3}}\]

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

\[{{625}^{\dfrac{1}{4}}}=\sqrt[4]{625}=\sqrt[4]{{{5}^{4}}}=5\]

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, \[{{\left( {{a}^{\dfrac{1}{n}}} \right)}^{m}}\]

**Note:**

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,

\[{{x}^{-1}}=\dfrac{1}{x}\]

\[\therefore {{\left( \dfrac{1}{16} \right)}^{\dfrac{-3}{4}}}={{(16)}^{\dfrac{3}{4}}}\]

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 \[8\]

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