If 20^{2-3x^2} = (40sqrt{5})^{3x^2-2},then x | vedclass

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(B) Given the equation: $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$.Note that $40\sqrt{5} = 20 	imes 2 	imes \sqrt{5} = 20 	imes \sqrt{4} 	imes \sqrt{5}

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$\pm \sqrt{\frac{2}{3}}$

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(B) Given the equation: $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$.Note that $40\sqrt{5} = 20 	imes 2 	imes \sqrt{5} = 20 	imes \sqrt{4} 	imes \sqrt{5} = 20 	imes \sqrt{20} = 20^{1} 	imes 20^{1/2} = 20^{3/2}$.Substituting this into the equation:$20^{2-3x^2} = (20^{3/2})^{3x^2-2}$$20^{2-3x^2} = 20^{\frac{3}{2}(3x^2-2)}$Since the bases are equal,we equate the exponents:$2-3x^2 = \frac{3}{2}(3x^2-2)$$2-3x^2 = -\frac{3}{2}(2-3x^2)$$(2-3x^2) + \frac{3}{2}(2-3x^2) = 0$$(2-3x^2)(1 + \frac{3}{2}) = 0$$(2-3x^2)(\frac{5}{2}) = 0$Since $\frac{5}{2} 
eq 0$,we must have $2-3x^2 = 0$.$3x^2 = 2$$x^2 = \frac{2}{3}$$x = \pm \sqrt{\frac{2}{3}}$.

If $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$,then $x$ is equal to

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