Question

Find the multiplicative inverse of ${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}}$.

Answer 1

Hint: Multiplicative inverse means the reciprocal of the given fraction. Reciprocal means writing the numerator in denominator and writing denominator in numerator. First of all, we will be simplifying the power of the fraction and then write its reciprocal.

Complete step-by-step answer:
In this question, we are given a fraction and we need to find its multiplicative inverse.
Multiplicative inverse simply means the reciprocal of the given fraction.
Reciprocal means writing the denominator of the fraction in numerator and writing the numerator in denominator.
Given fraction: ${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}}$
So, we need to find it’s reciprocal.
First of all, we need to make the power positive. For that we will take the reciprocal of the given fraction.
${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} = {\left( {\dfrac{{16}}{{81}}} \right)^{\dfrac{3}{4}}}$
Now, we can write 16 as 4 square and 81 as 9 square. Therefore,
${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} = {\left( {\dfrac{{16}}{{81}}} \right)^{\dfrac{3}{4}}} = {\left( {\dfrac{{{4^2}}}{{{9^2}}}} \right)^{\dfrac{3}{4}}} = {\left( {\dfrac{4}{9}} \right)^{\dfrac{3}{4} \times 2}} = {\left( {\dfrac{4}{9}} \right)^{\dfrac{3}{2}}}$
Now, take $\dfrac{1}{2}$ as separate power for both numerator and denominator.
${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} = {\left( {\dfrac{4}{9}} \right)^{\dfrac{3}{2}}} = {\left( {\dfrac{{{4^{\dfrac{1}{2}}}}}{{{9^{\dfrac{1}{2}}}}}} \right)^3}$
Now, $\dfrac{1}{2}$ power means square root. Therefore, square root of 4 is 2 and 9 is 3.
${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} = {\left( {\dfrac{{{4^{\dfrac{1}{2}}}}}{{{9^{\dfrac{1}{2}}}}}} \right)^3} = {\left( {\dfrac{2}{3}} \right)^3}$
Now, we simply need to find the cube of $\dfrac{2}{3}$.
${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} = {\left( {\dfrac{2}{3}} \right)^3} = \dfrac{{2 \times 2 \times 2}}{{3 \times 3 \times 3}} = \dfrac{8}{{27}}$
Now, we need to find its multiplicative inverse that is reciprocal.
${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} = \dfrac{{27}}{8}$
Hence, the multiplicative inverse of ${\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}}$ is $\dfrac{{27}}{8}$.

Note: Multiplicative inverse or reciprocal denoted by $\dfrac{1}{x}$ or ${x^{ - 1}}$ if x is the given number, is a number which when multiplied by x gives 1.
For example: In our question, we obtained $x = \dfrac{8}{{27}}$ and its inverse ${x^{ - 1}} = \dfrac{{27}}{8}$. So, when we multiply both this, we get
$\dfrac{8}{{27}} \times \dfrac{{27}}{8} = 1$