Question

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

Answer 1

This type of problem can be solved using the properties of power and division. First, we have to consider the given expression. We know that 16 is equal to ${{2}^{4}}$. Substitute in the given expression and use the property of division that is ${{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}$ to simplify the expression further. Then use the property of power that is ${{\left( {{a}^{n}} \right)}^{m}}={{a}^{n\times m}}$, in the denominator. Using the property of division $\left( \dfrac{1}{{{a}^{n}}} \right)={{a}^{-n}}$, we have to convert the fraction into an integer.

Complete step-by-step solution:
According to the question, we are asked to simplify ${{\left( \dfrac{1}{16} \right)}^{-\dfrac{3}{4}}}$.
We have been given the expression is ${{\left( \dfrac{1}{16} \right)}^{-\dfrac{3}{4}}}$. ---------(1)
Let us first consider the whole expression and use the property ${{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}$ in expression (1).
Therefore, we get
${{\left( \dfrac{1}{16} \right)}^{-\dfrac{3}{4}}}=\dfrac{1}{{{\left( 16 \right)}^{-\dfrac{3}{4}}}}$
But we know that ${{2}^{4}}=16$.
On substituting in the obtained expression, we get
$\Rightarrow {{\left( \dfrac{1}{16} \right)}^{-\dfrac{3}{4}}}=\dfrac{1}{{{\left( {{2}^{4}} \right)}^{-\dfrac{3}{4}}}}$ ----------(2)
Let us now consider the denominator of the simplified expression (2) that is ${{\left( {{2}^{4}} \right)}^{-\dfrac{3}{4}}}$.
We know that the power rule is ${{\left( {{a}^{n}} \right)}^{m}}={{a}^{n\times m}}$.
Using this rule in ${{\left( {{2}^{4}} \right)}^{-\text{ }\dfrac{3}{4}}}$, we get
${{\left( {{2}^{4}} \right)}^{-\text{ }\dfrac{3}{4}}}={{2}^{4\times -\text{ }\dfrac{3}{4}}}$
Here, in the power of 2, we find that 4 are common in both numerator and denominator.
Let us cancel 4 from the numerator and denominator.
$\Rightarrow {{\left( {{2}^{4}} \right)}^{-\text{ }\dfrac{3}{4}}}={{2}^{-3}}$
Substitute the simplified denominator in expression (2).
$\Rightarrow {{\left( \dfrac{1}{16} \right)}^{-\dfrac{3}{4}}}=\dfrac{1}{{{2}^{-3}}}$
We know that $\left( \dfrac{1}{{{a}^{n}}} \right)={{a}^{-n}}$. Using this property in the above expression, we get
$\Rightarrow {{\left( \dfrac{1}{16} \right)}^{-\dfrac{3}{4}}}={{2}^{-\left( -3 \right)}}$.
Since $-\left( -x \right)=x$, we get
${{\left( \dfrac{1}{16} \right)}^{-\dfrac{3}{4}}}={{2}^{3}}$
We know that the cube of 2 is equal to 8.
$\therefore {{\left( \dfrac{1}{16} \right)}^{-\dfrac{3}{4}}}=8$
Hence, the simplified form of ${{\left( \dfrac{1}{16} \right)}^{-\dfrac{3}{4}}}$ is 8.

Note: Whenever we get such types of problems, we should always compare the numerator and denominator with the power. Look for the common terms or try to convert the given constant to square or cube of any terms. Avoid calculation mistakes based on sign conventions. Do not neglect the negative sign in the power.