Question

If x = [ (2/3)2 ]3 × (1/3)-2 × 3-1 ×1/6 find the reciprocal of x.​

Answer 1

x = [(2/3)2]3 * (1/3) - 2 * 3 - 1 * 1/6

x = [4/3]3 * 1/3 - 2 * 2 * 1/6

x = [4/3] * 1 - 4 * 1/6

x = 4/3 - 4/6

x = 4/3 - 2/3

Hence, x = 2/3

Reciprocal of x is 1/x, so reciprocal becomes 3/2.

Answer 2

Answer:

$\frac{729}{32}$ is the required value of the reciprocal of x.

Step-by-step explanation:

Explanation:

Given in the question, x =  $[(\frac{2}{3})^2 ]^3 . (\frac{1}{3})^{-2} . 3^{-1} . \frac{1}{6}$

• BODMAS rule -The Bodmas rule is arranged according to the letters in the acronym BODMAS, which stand for brackets, order of powers or roots, division, and multiplication, A stands for addition, and S for subtraction.

• The BODMAS rule states that the bracket should be solved first. If there is no bracket, multiplication or division will take precedence and if it comes first in the mathematical phrase read from left to right.

Step 1:

We have,x =  $[(\frac{2}{3})^2 ]^3 . (\frac{1}{3})^{-2} . 3^{-1} . \frac{1}{6}$

First, we solve the bracket parts,

⇒x = $(\frac{2}{3})^6 . (\frac{1}{3} )^{-2}. (\frac{1}{3} ) . \frac{1}{6}$

⇒ x = $(2)^6 .( \frac{1}{3}) ^6 .(\frac{1}{3}) ^{-2} .(\frac{1}{3} ) . \frac{1}{3}.\frac{1}{2}$

⇒ x = $2^6$ . $(\frac{1}{3}) ^{6 + (-2)+1 + 1} .\frac{1}{2}$

⇒x = $2^6$ . $(\frac{1}{3}) ^6$ . $(2)^{-1}$

⇒ x= $2^{6 - 1} . (\frac{1}{3})^6$ = $\frac{2^5}{3^6}$

Now , from the question reciprocal of x = $\frac{3^6}{2^5}$ = $\frac{729}{32}$

Final answer:

Hence, $\frac{729}{32}$ is the reciprocal of x.

#SPJ2