Answer

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**Hint:**In order to solve this problem, we need to make the base common as we can see that the unknown is in the power. By making the base common we can directly equate the powers and solve for x.

**Complete step-by-step solution:**We are given expression with x as the unknown.

We need to find the value of x.

As we can see in the expression there are two different bases in the left-hand side and the right-hand side.

As our unknown is in the power, we need to make the base common in order to equate the powers.

Therefore, let's make the power to 20 on both sides.

The expression looks as follows,

${{20}^{3-2{{x}^{2}}}}={{\left( 40\sqrt{5} \right)}^{3{{x}^{2}}-2}}$

Let's take the 40 from the right-hand side inside the bracket.

For that we need to square the 40, the square of 40 is 1600.

Hence, substituting we get,

${{20}^{3-2{{x}^{2}}}}={{\left( \sqrt{1600\times 5} \right)}^{3{{x}^{2}}-2}}$

Solving we get,

${{20}^{3-2{{x}^{2}}}}={{\left( \sqrt{8000} \right)}^{3{{x}^{2}}-2}}$

But now we can see that 8000 is the cube of 20.

So we can replace 8000 by ${{20}^{3}}$, and we can replace the square root by $\dfrac{1}{2}$ in the power.

Substituting we get,

${{20}^{3-2{{x}^{2}}}}={{20}^{\dfrac{3}{2}\left( 3{{x}^{2}}-2 \right)}}$

Now, we can directly equate the powers with each other as base are the same on both sides.

Hence, equating the powers we get,

$3-2{{x}^{2}}=\dfrac{3}{2}\left( 3{{x}^{2}}-2 \right)$

Solving this for x we get,

$\begin{align}

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

& 6-4{{x}^{2}}=9{{x}^{2}}-6 \\

& 13{{x}^{2}}=12 \\

& {{x}^{2}}=\dfrac{12}{13} \\

\end{align}$

Taking the square root we get,

$x=\pm \sqrt{\dfrac{12}{13}}$

**Hence, the correct option is (b).**

**Note:**We can also do this question with a different approach. We can take the log on both sides and arrive at the same equation. It can be shown as follows,

Taking log on both sides we get,

$\ln \left( {{20}^{3-2{{x}^{2}}}} \right)=\ln \left( {{20}^{\dfrac{3}{2}\left( 3{{x}^{2}}-2 \right)}} \right)$

Now, we can use the property that $\ln \left( {{a}^{b}} \right)=a\ln b$ , we get,

$\begin{align}

& \left( 3-2{{x}^{2}} \right)\ln \left( 20 \right)=\left( \dfrac{3}{2}\left( 3{{x}^{2}}-2 \right) \right)\ln \left( 20 \right) \\

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

\end{align}$

Hence, we can see that we arrive at the same position.

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