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

Last updated date: 16th Jul 2024
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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
${{\left( \dfrac{1}{16} \right)}^{\dfrac{-3}{4}}}={{2}^{3}}\,\,=\,8$

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, ${{5}^{4}}=625$
${{625}^{\dfrac{1}{4}}}=\sqrt[4]{625}=\sqrt[4]{{{5}^{4}}}=5$
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 $8$