If (frac{2}{5})^{5} times (frac{25}{4})^{3} = | vedclass

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(D) Given equation: $(\frac{2}{5})^{5} 	imes (\frac{25}{4})^{3} = (\frac{5}{2})^{3x-2}$First,express all bases in terms of $(\frac{5}{2})$:$(\frac{2}

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$1$

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(D) Given equation: $(\frac{2}{5})^{5} 	imes (\frac{25}{4})^{3} = (\frac{5}{2})^{3x-2}$First,express all bases in terms of $(\frac{5}{2})$:$(\frac{2}{5})^{5} = (\frac{5}{2})^{-5}$$(\frac{25}{4})^{3} = ((\frac{5}{2})^{2})^{3} = (\frac{5}{2})^{6}$Now substitute these into the equation:$(\frac{5}{2})^{-5} 	imes (\frac{5}{2})^{6} = (\frac{5}{2})^{3x-2}$Using the law of exponents $a^m 	imes a^n = a^{m+n}$:$(\frac{5}{2})^{-5+6} = (\frac{5}{2})^{3x-2}$$(\frac{5}{2})^{1} = (\frac{5}{2})^{3x-2}$Since the bases are equal,equate the exponents:$1 = 3x - 2$$3x = 3$$x = 1$

If $(\frac{2}{5})^{5} \times (\frac{25}{4})^{3} = (\frac{5}{2})^{3x-2}$,then find $x$.

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