यदि 20^{2-3x^2} = (40sqrt{5})^{3x^2-2} है,तो | vedclass

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(B) दिया गया समीकरण: $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$ है।यहाँ $40\sqrt{5} = 20 	imes 2 	imes \sqrt{5} = 20 	imes \sqrt{20} = 20^{3/2}$ होता है

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) दिया गया समीकरण: $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$ है।यहाँ $40\sqrt{5} = 20 	imes 2 	imes \sqrt{5} = 20 	imes \sqrt{20} = 20^{3/2}$ होता है।अतः,$20^{2-3x^2} = (20^{3/2})^{3x^2-2} = 20^{\frac{3}{2}(3x^2-2)}$।घातांकों की तुलना करने पर:$2-3x^2 = \frac{3}{2}(3x^2-2)$$2-3x^2 = -\frac{3}{2}(2-3x^2)$$(2-3x^2)(1 + \frac{3}{2}) = 0$$2-3x^2 = 0$$3x^2 = 2$$x^2 = \frac{2}{3}$$x = \pm \sqrt{\frac{2}{3}}$।

यदि $20^{2-3x^2} = (40\sqrt{5})^{3x^2-2}$ है,तो $x$ का मान ज्ञात कीजिए।

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