Question

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Answer 1

Given & Find :-

$\rm If \: x = { \huge[}{ \big( }\dfrac{2}{3} \big) ^{2} { \huge]} ^{3} \times {\huge(} \dfrac{1}{3}{ \huge)}^{ - 2} \times 3 ^{ - 1} \times \dfrac{1}{6} , \: find \: the \: reciprocal \: of \: x \:$

We need to know

For solving this we need to know Exponents . Only , here we need some of them so, it is discussed below :-

1. If there are two powers written see in first fraction there are two powers written which include 2 and 3 . So, in this case we will have to multiply both the powers.
2. If the power is negative , see in the second fraction the power is negative so, in this case we will have to reverse the fraction means reciprocal or flip the fraction because it will help us to calculate.

These were the two cases we need to know . For solving this .

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So, let's solve :-

$\ { \huge[}{ \big( }\dfrac{2}{3} \big) ^{2} { \huge]} ^{3} \times {\huge(} \dfrac{1}{3}{ \huge)}^{ - 2} \times 3 ^{ - 1} \times \dfrac{1}{6}$

Firstly removing the third bracket or box bracket and reversing the fraction and making the power positive.

$\rm{ \big( }\dfrac{2}{3} \big) ^{6} \times {3}^{2} \times { \huge(}\dfrac{1}{3} {\huge)} ^{1} \times \dfrac{1}{6}$

Now, simplifying the fraction

$\rm\dfrac{64}{729} \times 9 \times \dfrac{1}{3} \times \dfrac{1}{6}$

Now multiplying the last two fractions together

$\rm\dfrac{64}{729} \times { \cancel9 }\times \dfrac {1}{\cancel{18}}$

$\longmapsto\rm\dfrac{ \cancel{64}}{729} \times \dfrac{1}{ \cancel{ 2}}$

$\longmapsto\rm \: \: \bf \dfrac{ {32}}{729}$

Now, as it was said that to find the reciprocal of X . So, we have already found x but let's now reciprocal of x .

$\rm \: x = \bf\dfrac{ {729}}{32} \: \: \: { \huge\star}$

$\therefore \rm \: Value \: of \: X \: is \:{\bf \dfrac { {32}}{729}}\:and\:reciprocal\:of\:X \:is\:\bf\:\dfrac{729}{32}$

Answer 2

Given

• x = $[(\dfrac{2}{3})^{2}]^{3} \times (\dfrac{1}{3})^{-2} \times 3^{-1} \times \dfrac{1}{6}$

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To Find

• Reciprocal of 'x'

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Solution

So first we need to apply the laws of exponents and simplify the value of 'x'.

Law of Exponents to be Used ⇒

• $(a^{m})^{n} = a^{m\times n}$

• $(a)^{-m} = \dfrac{1}{a^{m}}$

$\implies [(\dfrac{2}{3})^{2}]^{3} \times (\dfrac{1}{3})^{-2} \times 3^{-1} \times \dfrac{1}{6}$

$\implies [(\dfrac{2}{3})^{2\times 3 }] \times (3)^{2} \times \dfrac{1}{3} \times \dfrac{1}{6}$

$\implies [(\dfrac{2}{3})^{6}] \times 9 \times \dfrac{1}{18}$

$\implies \dfrac{64}{729} \times \dfrac{1}{2}$

$\implies \dfrac{32}{729}$

$\bf \therefore x = \dfrac{32}{729}$

The product of a number and its reciprocal is always one, so the reciprocal of 'x' would be,

Let reciprocal of 'x' be 'y'.

$\implies y\times \dfrac{32}{729} = 1$

$\implies y = \dfrac{729}{32}$

$\bf \therefore \ The \ reciprocal \ of \ x \ is \ \dfrac{729}{32}.$

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