Answer

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**Hint:**To solve the given expression we will use the properties of exponentiation and logarithm. First we will convert the given expression in exponentiation form by using the relation that ${{\log }_{a}}x=y$ is equivalent to ${{a}^{y}}=x$. By using this property we form an equation and simplify it to get the desired answer.

**Complete step-by-step solution:**

We have been given that ${{\log }_{x}}\left( \dfrac{1}{8} \right)=-\dfrac{3}{2}$.

We have to solve the given expression and find the value of x.

We know that logarithm is the inverse function to the exponentiation. The exponent of a number says how many times to use a number in multiplication. We will use the basic properties of logarithm to solve further.

We know that ${{\log }_{a}}x=y$ is equal to ${{a}^{y}}=x$.

So by applying the above property we will get

$\begin{align}

& \Rightarrow {{\log }_{x}}\left( \dfrac{1}{8} \right)=-\dfrac{3}{2} \\

& \Rightarrow {{x}^{\dfrac{-3}{2}}}=\dfrac{1}{8} \\

\end{align}$

Now, we know that ${{x}^{-a}}=\dfrac{1}{{{x}^{a}}}$

Applying the property to the above obtained equation we will get

\[\Rightarrow \dfrac{1}{{{x}^{\dfrac{3}{2}}}}=\dfrac{1}{8}\]

Now, simplifying the above equation we will get

$\Rightarrow {{x}^{\dfrac{3}{2}}}=8$

Now, we know that $8={{2}^{3}}$

Substituting the value we will get

$\Rightarrow {{x}^{\dfrac{3}{2}}}={{2}^{3}}$

Now, taking the square on both sides we will get

\[\Rightarrow {{\left( {{x}^{\dfrac{3}{2}}} \right)}^{2}}={{\left( {{2}^{3}} \right)}^{2}}\]

Now, we know that ${{\left( {{x}^{m}} \right)}^{n}}={{x}^{mn}}$

Applying the property to the above obtained equation we will get

\[\begin{align}

& \Rightarrow \left( {{x}^{\dfrac{3}{2}\times 2}} \right)=\left( {{2}^{3\times 2}} \right) \\

& \Rightarrow {{x}^{3}}=\left( {{2}^{3\times 2}} \right) \\

\end{align}\]

Now, taking the cube root both the sides we will get

$\begin{align}

& \Rightarrow x={{2}^{2}} \\

& \Rightarrow x=4 \\

\end{align}$

**So on solving the given expression we get the value $x=4$.**

**Note:**To solve such types of questions students must have the knowledge of properties of logarithm and properties of exponentiation. The possibility of mistake is while applying the property ${{\log }_{a}}x=y$. Students must carefully compare the values of a, x and y and substitute properly.

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