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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the equation: $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$.Notice that $40\sqrt{5} = 20 	imes 2\sqrt{5} = 20 	imes \sqrt{4 	imes 5} = 20 	imes \sqrt{20} = 20^1 	imes 20^{1/2} = 20^{3/2}$.Substituting this into the equation,we get: $20^{2-3x^2} = (20^{3/2})^{3x^2-2}$.Equating the exponents: $2-3x^2 = \frac{3}{2}(3x^2-2)$.Let $y = 3x^2-2$. Then the equation becomes $-y = \frac{3}{2}y$.This implies $y(1 + \frac{3}{2}) = 0$,so $y = 0$.Thus,$3x^2-2 = 0$,which gives $3x^2 = 2$.Therefore,$x^2 = \frac{2}{3}$,so $x = \pm \sqrt{\frac{2}{3}}$.

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

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