On the set of all non-zero reals,an operation | vedclass

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(C) Given binary operation is $a * b = \frac{3ab}{2}$.First,we find the identity element $e$ such that $a * e = a$.$\frac{3ae}{2} = a \implies e = \fr

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$1/6$

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(C) Given binary operation is $a * b = \frac{3ab}{2}$.First,we find the identity element $e$ such that $a * e = a$.$\frac{3ae}{2} = a \implies e = \frac{2}{3}$.Now,we find the inverse $3^{-1}$ such that $3 * 3^{-1} = e = \frac{2}{3}$.$\frac{3 \cdot 3 \cdot 3^{-1}}{2} = \frac{2}{3} \implies \frac{9 \cdot 3^{-1}}{2} = \frac{2}{3} \implies 3^{-1} = \frac{4}{27}$.Also,$2 * x = \frac{3 \cdot 2 \cdot x}{2} = 3x$.Given equation: $(2 * x) * 3^{-1} = 4^{-1}$.Note that $4^{-1}$ is the inverse of $4$,so $4 * 4^{-1} = \frac{2}{3} \implies \frac{3 \cdot 4 \cdot 4^{-1}}{2} = \frac{2}{3} \implies 6 \cdot 4^{-1} = \frac{2}{3} \implies 4^{-1} = \frac{1}{9}$.Substituting these into the equation: $(3x) * \frac{4}{27} = \frac{1}{9}$.$\frac{3 \cdot (3x) \cdot (4/27)}{2} = \frac{1}{9}$.$\frac{36x}{54} = \frac{1}{9} \implies \frac{2x}{3} = \frac{1}{9}$.$x = \frac{3}{18} = \frac{1}{6}$.

$^*$	$1$	$2$	$3$	$4$	$5$
$1$	$1$	$1$	$1$	$1$	$1$
$2$	$1$	$2$	$2$	$2$	$2$
$3$	$1$	$2$	$3$	$3$	$3$
$4$	$1$	$2$	$3$	$4$	$4$
$5$	$1$	$2$	$3$	$4$	$5$

On the set of all non-zero reals,an operation $$ is defined as $a b = \frac{3ab}{2}$. In this group,a solution of $(2 * x) * 3^{-1} = 4^{-1}$ is

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On the set of all non-zero reals,an operation $*$ is defined as $a * b = \frac{3ab}{2}$. In this group,a solution of $(2 * x) * 3^{-1} = 4^{-1}$ is

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Let $^*$ be a binary operation on the set $Q$ of rational numbers defined as $a \,^* \,b = a^{2} + b^{2}$. Which of the following is true?

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