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On the set of all non-zero reals, an operation * is defined as $a^{*} b=\frac{3 a b}{2}$. In this group, a solution of $\left(2^{*} x\right) * 3^{-1}=4^{-1}$ is
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The correct answer is:
$1 / 6$
Given binary operation is
$$
a^{*} b=\frac{3 a b}{2}
$$
Now, $\quad 2 * x=\frac{3}{2} \cdot 2 x=3 x[$ from Eq. (i) $] \ldots$ (ii) and $(3 x)^{*} \frac{1}{3}=\frac{3}{2} \cdot 3 x \cdot \frac{1}{3}=\frac{3 x}{2}$
Then, $\left(2^{*} x\right)^{*} 3^{-1}=4^{-1}$
$\begin{aligned} \Rightarrow &(3 x)^{\frac{*}{n}} \frac{1}{3}=\frac{1}{4} & \text { [from Eq. (ii)] } \\ \Rightarrow & \frac{3 x}{2} &=\frac{1}{4} & \text { [from Eq. (iii)] } \\ \Rightarrow & x &=\frac{1}{6} & \end{aligned}$
$$
a^{*} b=\frac{3 a b}{2}
$$
Now, $\quad 2 * x=\frac{3}{2} \cdot 2 x=3 x[$ from Eq. (i) $] \ldots$ (ii) and $(3 x)^{*} \frac{1}{3}=\frac{3}{2} \cdot 3 x \cdot \frac{1}{3}=\frac{3 x}{2}$
Then, $\left(2^{*} x\right)^{*} 3^{-1}=4^{-1}$
$\begin{aligned} \Rightarrow &(3 x)^{\frac{*}{n}} \frac{1}{3}=\frac{1}{4} & \text { [from Eq. (ii)] } \\ \Rightarrow & \frac{3 x}{2} &=\frac{1}{4} & \text { [from Eq. (iii)] } \\ \Rightarrow & x &=\frac{1}{6} & \end{aligned}$
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